Limit of sequences, compl Lecture 7

Chapter 1 Infinite SeriesGenerally, for the given sequence ,.......,......,3,21n a a a a the expression formed by the sequence ,.......,......,3,21n a a a a .......,.....321+++++n a a a ais called the infinite series of the constants term, denoted by∑∞=1n na, that is∑∞=1n na=.......,.....321+++++n a a a aWhere the nth term is said to be the general term of the series, moreover, the nth partial sum of the series is given by=n S ......321n a a a a ++++1.1 Determine whether the infinite series converges or diverges.While it ’s possible to add two numbers, three numbers, a hundred numbers, or even a million numbers, it ’s impossible to add an infinite number of numbers.To form an infinite series we begin with an infinite sequence of real numbers: .....,,,3210a a a a , we can not form the sum of all the k a (there is an infinite number of the term), but we can form the partial sums∑===0000k k a a S∑==+=1101k k a a a S∑==++=22102k k a a a a S∑==+++=332103k k a a a a a S……………….∑==+++++=nk k n n a a a a a a S 03210.......Definition 1.1.1If the sequence {n S } of partial sums has a finite limit L, We write ∑∞==0k k a Land say that the series ∑∞=0k kaconverges to L. we call L thesum of the series.If the limit of the sequence {n S } of partial sums don ’t exists, we say that the series∑∞=0k kadiverges.Remark it is important to note that the sum of a series is not a sum in the ordering sense. It is a limit.EX 1.1.1 prove the following proposition: Proposition1.1.1: (1) If 1<x then the∑∞=0k k a converges, and;110xx k k -=∑∞=(2)If ,1≥x then the∑∞=0k kxdiverges.Proof: the nth partial sum of the geometric series∑∞=0k katakes the form 1321.......1-+++++=n n x x x x S ① Multiplication by x gives).......1(1321-+++++=n n x x x x x xS =n n x x x x x +++++-1321.......Subtracting the second equation from the first, we find thatn n x S x -=-1)1(. For ,1≠x this givesxx S nn --=11 ③If ,1<x then 0→n x ,and this by equation ③.x xx S n n n n -=--=→→1111lim lim 00 This proves (1).Now let us prove (2). For x=1, we use equation ① and device that ,n S n =Obviously,∞=∞→nn Slim ,∑∞=0k kadiverges.For x=-1 we use equation ① and we deduce If n is odd, then 0=n S , If n is even, then .1-=n SThe sequence of partial sum n S like this 0,-1,0,-1,0,-1………..Because the limit of sequence }{n S of partial sum does not exist. By definition 1.1.1, we have the series ∑∞=0k Kxdiverges.(x=-1).For 1≠x with ,1>x we use equation ③. Since in thisinstance, we have -∞=--=∞→∞→xx S nn n n 11lim lim . The limit of sequence of partial sum not exist, the series∑∞=0k kxdiverges.Remark the above series is called the geometric series. It arises in so many different contexts that it merits special attention.A geometric series is one of the few series where we can actually give an explicit formula for n S ; a collapsing series is another.Ex.1.1.2 Determine whether or not the series converges∑∞=++0)2)(1(1k k k Solution in order to determine whether or not this series converges we must examine the partial sum. Since2111)2)(1(1+-+=++k k k kWe use partial fraction decomposition to write2111111........................41313121211)2111()111(..............)4131()3121()2111()2)(1(1)1(1..............3.212.11+-+++-++-+-+-=+-+++-++-+-+-=++++++⨯+⨯=n n n n n n n n n n n n S n Since all but the first and last occur in pairs with opposite signs, the sum collapses to give 211+-=n S n Obviously, as .1,→∞→n S n this means that the seriesconverges to 1. 1)211(lim lim =+-=∞→∞→n S n n n therefore 1)2)(1(10=++∑∞=n k k EX.1.1.3 proves the following theorem:Theorem 1.1.1 the kth term of a convergent series tends to 0; namely if∑∞=0k kaConverges, by definition we have the limit of thesequence }{n S of partial sums exists. Namelyl a S nk k n n n ==∑=∞→∞→0lim limObviously.lim lim 01l a S nk k n n n ==∑=∞→-∞→since 1--=n n s s a n ,we have0lim lim )(lim lim 11=-=-=-=-∞→∞→-∞→∞→l l S S S S a n n n n n n n n nA change in notation gives 0lim =∞→n k a .The next result is an obviously, but important, consequence of theorem1.1.1. Theorem 1.1.2 (A diverges test) if 0lim ≠∞→k k a , or ifn k a ∞→lim does not exist, then the series ∑∞=0k k a diverges.Caution, theorem 1.1.1 does not say that if 0lim =∞→k k a , and then∑∞=0k kaconverge. In fact, there are divergent series forwhich 0lim =∞→k k a .For example, theseries .....1. (2)11111++++=∑∞=nkk . Since it is sequence }{n S ofpartial sum nn n n S n =>+++=1 (2)111}{ is unbounded. So∞===∞→∞→n S n n n lim lim , therefore the series diverges.But 01lim lim ===∞→∞→ka k k kEX.1.1.3 determine whether or not the series: (54)433221010+++++=+∑∞=k k k Converges. Solution since 01111lim 1limlim ≠=+=+==∞→∞→∞→kk k a k k k k , this seriesdiverges.EX.1.1.4 Determine whether or not the series∑∞=021k kSolution 1 the given series is a geometric series.121,)21(00<==∑∑∞=∞=x and xk k k k, by proposition 1.1.1 we know thatseries converges.Solution 2 ,21 (412111)-++++=n n S ① ,2121.........21212121132n n n S +++++=-②①-② (1-21))211(2,211n n n n S S -=-=.2)211(2lim lim =-=∞→∞→n n n n SBy definition of converges of series, this series converges.EX.1.1.5 proofs the following theorem: Theorem 1.1.2 If the series ∑∑∞=∞=0k kk k band a converges, then(1))(0∑∞=+k k kb aalso converges, and is equal the sum of the twoseries.(2) If C is a real number, then ∑∞=0k kCaalso converges.Moreover ifl ak k=∑∞=0thenCl Cak k=∑∞=0.Proof let ∑∑====nk k nnk k nb S a S20)1(,∑∑===+=nk k nnk k k nCa S b a S40)3(,)(Note that )1()4()2()1()3(n n n n n CS S and S S S =+=Since (),lim ,lim )2(1m S l S n n n n ==∞→∞→ Then m l S S S S S n n n n n n n n n +=+=+=∞→∞→∞→∞→)2()1()2()1()3(lim lim )(lim lim .lim lim lim )1()1()4(Cl S C CS S n n n n n n ===∞→∞→∞→Theorem 1.1.4 (squeeze theorem)Suppose that }{}{n n c and a both converge to l and thatn n n c b a ≤≤ for ,k n ≥(k is a fixed integer), then }{n b alsoconverges to l .Ex.1.1.6 show that 0sin lim3=∞→nnn . Solution For,1≥n ,1)sin (13nn n n ≤≤- since,0)1(lim ,0)1(lim ==-∞→∞→n and nn n the result follows by the squeeze theorem.For sequence of variable sign, it is helpful to have the following result.EX1.1.7 prove that the following theorem holds.Theorem 1.1.5 If 0lim ,0lim==∞→∞→n n n n a then a , Proof since ,n n n a a a ≤≤- from the theorem 1.1.4 Namely the squeeze theorem, we know the result is true.Exercise 1.1(1) An expression of the form 123a a a +++…is called (2) A series 123a a a +++…is said to converge if the sequence{}S n converges, where S n =1. The geometric series 2a ar ar +++…converges if ; in this case the sum of the series is2. If lim 0n n a →∞≠, we can be sure that the series 1nn a∞==∑3. Evaluate 0(1),02k k r r r ∞=-<<∑.4. Evaluate 0(1),11k k k x x ∞=--<<∑.5. Show that 1ln1k kk ∞=+∑diverges. Find the sums of the series 6-116. 31(1)(2)k k k ∞=++∑ 7.112(1)k k k ∞=+∑ 8.11(3)k k k ∞=+∑ 9.0310k k ∞=∑10.0345k k k k ∞=+∑ 11.3023k k k +∞=∑12. Derive the following results from the geometric series 2201(1),||11k k k x x x ∞=-=<+∑. Test the following series for convergence:13. 11n n n ∞=+∑ 14.3012k k ∞+=∑1.2 Series With Positive Terms1.2.1 The comparison TestThroughout this section, we shall assume that our numbers n a are x 0≥, then the partial sum 12n n S a a a =+++… are increasing, i.e. 1231n n S S S S S +≤≤≤≤≤≤……If they are to approach a limit at all, they cannot become arbitrarily large. Thus in that case there is a number B such that n S B ≤ for all n. Such a number B is called an upperbound. By a least upper bound we mean a number S which is an upper bound, and such that every upper bound B is S ≥. We take for granted that a least upper bound exists. The collection of numbers {}n S has therefore a least upper bound, i.e., there is a smallest numbers such that n S S ≤ for all n. In that case, the partial sums n S approach S as a limit. In other words, given any positive number 0ε>, we have n S S S ε-≤≤ for all n sufficiently large.This simply expresses the fact S is the least of all upper bounds for our collection of numbers n S . We express this as a theorem.Theorem 1.2.1 Let {}(1,2,n a n =…) be a sequence of numbers 0≥and let 12n n S a a a =+++…. If the sequence of numbers {}n S is bounded, then it approaches a limit S , which is its least upper bound.Theorem 1.2.2 A series with nonnegative terms converges if and only if the sequence of partial sums is bounded above.Theorem 1.2.1 and 1.2.2 give us a very useful criterion todetermine when a series with positive terms converges.S 1 S 2 S n SThe convergence or divergence of a series with nonnegative terms is usually deduced by comparison with a series of known behavior.Theorem 1.2.3(The Ordinary Comparison Test) Let1nn a∞=∑and1nn b∞=∑be two series, with 0n a ≥ for all n and 0n b ≥ forall n. Assume that there is a numbers 0c >, such that n n a cb ≤ for all n, and that1nn b∞=∑ converges, then1nn a∞=∑converges, and11nn n n ac b ∞∞==≤∑∑.Proof: We have1212121()n n n n n a a a cb cb cb c b b b c b ∞=+++≤+++=+++≤∑……….This means that 1n n c b ∞=∑ is a bound for the partialsums 12n a a a +++….The least upper bound of these sums is therefore 1n n c b ∞=≤∑, thusproving our theorem.Theorem 1.2.3 has an analogue to show that a series does not converge.Theorem 1.2.4(Ordinary Comparison Test) Let1nn a∞=∑ and1nn b∞=∑ be two series, with n a and 0n b ≥ for all n. Assume thatthere is a number 0c > such that n n a cb ≥ for all n sufficiently large, and1nn b∞=∑ does not converge, then1nn a∞=∑ diverges. Proof. Assume n n a cb ≥for 0n n ≥, since 1nn b∞=∑diverges, wecan make the partial sum0001Nnn n N n n bb b b +==+++∑…arbitrarily largeas N becomes arbitrarilylarge. ButNNNnnnn n n n n n a cbc b ===≥=∑∑∑.Hencethepartialsum121NnN n aa a a ==+++∑… are arbitrarily large as N becomes arbitrarilylarge, are hence1nn a∞=∑ diverges, as was to be shown.Remark on notation you have easily seen that for each 0j ≥,kk a∞=∑ converges iff1kk j a∞=+∑ converges. This tells us that, indetermining whether or not a series converges, it does not matter where we begin the summation, where detailed indexing would contribute nothing, we will omit it and write∑without specifying where the summation begins. Forinstance, it makes sense to you that21k ∑ converges and1k ∑diverges without specifying where we begin the summation. But in the convergent case it does, however, affect the sum.Thus for example0122kk ∞==∑, 1112kk ∞==∑, 21122kk ∞==∑, and so forth. Ex 1.2.1 Prove that the series211n n ∞=∑ converges. Solution Let us look at the series:22222222211111111112345781516+++++++++++………We look at the groups of terms as indicated. In each group of terms, if we decrease the denominator in each term, then we increase the fraction. We replace 3 by 2 , then 4,5,6,7 by 4, then we replace the numbers from 8 to 15 by 8, and so forth. Ourpartial sums therefore less than or equal to222222221111111112244488++++++++++……… and we note that 2 occurs twice, 4 occurs four times, 8 occurs eight times, and so forth. Our partial sum are therefore less than or equal to222222221111111112244488++++++++++……… and we note that 2 occurs twice, 4 occurs four times, 8 occurs eight times, and so forth. Hence the partial sums are less than or equal to2222124811124848+++++++1…=1+?2 Thus our partial sums are less than or equal to those of the geometric series and are bounded. Hence our seriesconverges.Generally we have the following result: The series1111111234p p p p p n n n ∞==++++++∑……, where p is a constant, is called a p-series.Proposition1.2.1. If 1p >, the p-series converges; and if1p ≤, then the p-series diverges.Ex 1.2.2 Determine whether the series 2311n n n ∞=+∑ converges.Solution We write 2323111(1)1111n n n n n n ==++++. Then we see that 23111122n n n n≥=+. Since11n n ∞=∑ does not converge, it follows that the series 2311n n n ∞=+∑ does not converge either. Namely thisseries diverges.Ex 1.2.3 Prove the series 241723n n n n ∞=+-+∑ converges.Proof :Indeed we can write2222424334477(1)171331123(2())2()n n n n n n n n n n n n+++==-+-+-+ For n sufficiently large, the factor 23471312()n n n+-+ is certainly bounded, and in fact is near 1/2. Hence we can compare ourseries with21n ∑ to see converges, because ∑21n convergesand the factor is bounded.Ex.1.2.5 Show that1ln()k b +∑ diverges.Solution 1 We know that as k →∞,ln 0kk→. It follows that ln()0k b k b +→+, and thus that ln()ln()0k b k b k bk k b k+++=→+. Thus for k sufficiently large, ln()k b k +< and 11ln()k k b <+. Since 1k ∑diverges, we can conclude that1ln()k b +∑ diverges.Solution 2 Another way to show that ln()k b k +< for sufficiently large k is to examine the function ()ln()f x x x b =-+. At 3x = the function is positive:(3)3ln93 2.1970f =-=->Since '1()10f x x b=->+ for all 0x >, ()0f x > for all 3x >. It follows thatln()x b x +< for all 3x ≥.We come now to a somewhat more comparison theorem. Our proof relies on the basic comparison theorem.Theorem 1.2.5(The Limit Comparison Test) Letka∑ andkb∑ be series with positive terms. If lim()k k ka lb →∞=, where l issome positive number, then ka∑ andkb∑converge ordiverge together.Proof Choose ε between 0 and l , since k ka lb →, weknow for all k sufficiently large (for all k greater than some0k ) ||kka lb ε-<. For suchkwe have kka l lb εε-<<+, and thus()()k k k l b a l b εε-<<+ this last inequality is what we needed.(1) I fka∑converges, then()kl b ε-∑converges, and thuskb∑converges.(2) I fkb∑converges, then()kl bε+∑converges, and thuska∑converges.To apply the limit comparison theorem to a series k a ∑, we must first find a serieskb∑of known behavior for whichkka b converges to a positive number. Ex 1.2.6 Determine whether the series sinkπ∑convergesor diverges.Solution Recall that as sin 0,1x x x →→. As ,0k kπ→∞→ and thussin 1k kππ→. Sincek π∑diverges, sosin()k π∑diverges.Ex 1.2.7 Determine whether theseriesconverges or diverges.Solution For large value of k, dominates thenumeratorand 2k dominates the denominator, thus, forsuch k,252k=. Since22512k÷==→And2255122k k=∑∑converges, this series converges.Theorem 1.2.6 Letka∑and k b∑be series with positive terms and suppose thus 0kkab→, then(1)I fkb∑converges, then k a∑converges.(2)I fka∑diverges, then k b∑diverges.(3)I fka∑converges, then k b∑may converge or diverge.(4)I fkb∑diverges, then k a∑may converge or diverge.[Parts (3) and (4) explain why we stipulated 0l>in theorem1.2.5]1.2.2 The root test and the ratio testTheorem 1.2.7 (the root test, Cauchy test) let ∑k a be a series with nonnegative terms and suppose thatρ==∞→∞→kkkkkkaa1limlim, if ρ<1, ∑k a converges, if ρ>1, ∑k a diverges, if ρ=1, the test is inconclusive.Proof we suppose first ρ<1 and choose μso that1<<u ρ. Since ρ→kk a 1)(, we have μ<k ka 1, for all k sufficientlylarge thus k k a μ< for all k sufficiently large since∑kμconverges (a geometric series with 0<1<μ), we know by theorem 1.2.5 that∑kaconverges.We suppose now that 1>ρand choose μso that 1>>u ρ. since ρ→kk a 1)(, we have μ>kk a 1)( for all k sufficiently large. Thus k k a μ> for all k sufficiently large.Since∑kμdiverges (a geometric series with 1>μ ) thetheorem 1.2.6 tell us that∑kadiverges.To see the inconclusiveness of the root test when 1=ρ, note that 1)(1→kk a for both:112∑∑k and k ,11)1()1()(221121=→==kk kk k ka 11)1()(11→==k k kk kk aThe first series converges, but the second diverges. EX.1.2.7 Determine whether the series ∑k k )(ln 1convergesor diverges.Solution For the series ∑kk )(ln 1, applying the root test we have0ln 1lim)(lim 1==∞→∞→ka k kk k , the series converges. EX.1.2.8 Determine whether series ∑3)(2k kconverges ordiverges.Solution For the series ∑k k)3(2, applying the root test, wehave1212]1[2)1(.2)(3331>=⨯→==k k kk k k a . So the series diverges.EX1.2.9 Determines whether the series kk∑-)11(convergesor diverges.Solution in the case of kk ∑-)11(, we have 111)(1→-=ka k k . Ifapplying the root test, it is inconclusive. But sincek k k a )11(-=converges to e1and not to 0, the series diverges.We continue to consider only series with terms 0≥. To compare such a series with a geometric series, the simplest test is given by the ratio test theoremTheorem 1.2.8 (The ratio test, DAlembert test) let ∑kabea series with positive terms and suppose thatλ=+∞→kk k a a 1lim, If ,1<λ∑kaconverges, if ,1>λ∑kadiverges.If the ,1=λthe test is inconclusive.Proof we suppose first that ,1<λ since 1lim1<=+∞→λkk k a a So there exists some integer N such that if n ≥NC a a nn ≤+1Then N N N N N a C Ca a Ca a 212,1≤≤≤+++ and in general by induction ,N k k N a C a ≤+Thusca c c c c a a c a c ca a aNk N Nk N N N k N Nn n-≤++++≤++++≤∑+=11)........1( (322)Thus in effect, we have compared our series with a geometric series, and we know that the partial sums are bounded. This implies that our series converges.The ratio test is usually used in the case of a series with positive terms n a such that .1lim 1<=+∞→λnn n a a EX.1.2.10 show that the series∑∞=13n nnconverges. Solution we let ,3n n na = then 31.13.3111n n n n a a n n n n +=+=++, this ratioapproaches ∞→n as 31, and hence the ratio test is applicable: the series converges.EX1.2.11 show that the series ∑!k k kdiverges.Solution we have kk kk n n kk k k k k k a a )11()1(!)!1()1(11+=+=++=++ So e ka a k k n n n =+=∞→+∞→)11(lim lim1 Since 1>e , the series diverges. EX.1.2.12 proves the series∑+121k diverges.Solution since kkk k k k a a k k 32123212112.1)1(211++=++=+++=+ 13212limlim 1=++=∞→+∞→kk a a k kk k . Therefore the ratio test is inconclusive. We have to look further. Comparison with the harmonic series shows that the series diverges:∑++=+>+)1(21,11.21)1(21121k k k k dverges. Exercise 1.21. The ordinary comparison test says that if ____ and if∑ib converges. Then ∑kaalso converges.2. Assume that 00>≥k k b and a . The limit comparison Test says that if 0<____<+∞ then ∑kaand∑kbconverges or divergetogether. 3. Let nn n a a 1lim+∞→=ρ. The ratio Test says that a series ∑kaofpositive terms converges if ___, diverges if ____and may do either if ___.Determine whether the series converges or diverges 4.∑+13k k5.∑+2)12(1k 6.∑+11k 7.∑-kk 2218. ∑+-1tan 21k k9.∑321k10. ∑-k )43( 11.∑k kln 12.∑!10k k13. ∑k k 114.∑k k 100!15.∑++k k k 6232 16.kk ∑)32( 17.∑+k 11.18.∑k k 410!19. Let }{n a be a sequence of positive number and assume that na a n n 111-≥+ for all n. show that the series ∑nadiverges.1.3 Alternatingseries,Absolute convergenceandconditional convergenceIn this section we consider series that have both positive and negative terms.1.3.1 Alternating series and the tests for convergence The series of the form .......4321+-+-u u u u is called the alternating series, where 0>n u for all n, here two example:∑∞=--=+-+-+-11)1(....61514131211n n n ,11)1( (65544332211)+-=+-+-+-∑∞=n n nWe see from these examples that the nth term of an alternating series is the form n n n n n n u a or u a )1()1(1-=-=-, where n u is a positive number (in fact n n a u =.)The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges.Theorem 1.3.1 (Leibniz Theorem)If the alternating seriesnn nu∑∞=-1)1(satisfy:(1) 1+≥n n u u (n=1,2………); (2) 0lim =∞→n n u ,then the series converges. Moreover, it is sum 1u s ≤, and the error n r make by using n s of the first n terms to approximate the sum s of the series is not more than 1+n u , that is, 1+≤n n u r namely 1+≤-=n n n u s s r .Before giving the proof let us look at figure 1.3.1 which gives a picture of the idea behind the proof. We first plot 11u s =on a number line.To find 2s we subtract 2u , so 2s is the left of 1s . Then to find3s we add 3u , so 3s is to the right of 2s . But, since 3u <2u , 3s is tothe left of 1s . Continuing in this manner, we see that the partial sums oscillate back and forth. Since 0→n u , the successive steps are becoming smaller and smaller. The even partial sums,........,,642s s s are increasing and the odd partial sums,........,,531s s s are decreasing. Thus it seems plausible that both areconverging to some number s, which is the sum of the series. Therefore, in the following proof we consider the even and odd partial sums separatelyWe give the following proof of the alternating series test. Wefirst consider the even partial sums: ,0212≥-=u u s Since 12u u ≤,)(24324s u u s s ≥-+= since u u ≤4In general, 22212222)(---≥-+=n n n n n s u u s s since 122-≤n n u u Thus .........................02642≤≤≤≤≤≤n s s s s But we can also writen n n n u u u u u u u u s 21222543212)(....)()(--------=--Every term in brackets is positive, so 12u s n ≤ for all n. therefore, the sequence }{2n s of even partial sums is increasing and bounded above. It is therefore convergent by the monotonic sequence theorem. Let ’s call it is limit s, that is, s s n n =∞→2limNow we compute the limit of the odd partial sums:scondition by s u s u s s n n n n n n n n n =+=+=+=+∞→∞→+∞→+∞→))2((0lim lim )(lim lim 12212212Since both the even and odd partial sums converge to s, we have s s n n =∞→lim , and so the series is convergent.EX.1.3.1 shows that the following alternating harmonic series is convergent:.)1( (41312111)1∑∞=--=+-+-n n n Solution the alternating harmonic series satisfies (1) nu n u n n 1111=<+=+; (2) 01lim lim ==∞→∞→n u n n n So the series is convergent by alternating series Test.Ex. 1.3.2 Test the series ∑∞=--1143)1(n n n nfor convergence anddivergence.Solution the given series is alternating but 043143lim 143limlim ≠=-=-=∞→∞→∞→nn n u n n n n So condition (2) is not satisfied. Instead, we look at the limit of the nth term of the series: 143)1(lim lim --=∞→∞→n na n n n This limit does not exist, so the series diverges by the test for divergence.EX.1.3.3 Test the series ∑∞=+-121)1(n nn for convergence ordivergence.Solution the given series is alternating so we try to verify conditions (1) and (2) of the alternating series test.Unlike the situation in example 1.3.1, it is not obvious the sequence given by 12+=n nu n is decreasing. If we consider the related function1)(2+=x xx f ，we easily find that 10)1(1)1(21)(22222222'><+-=+-+=x whenver x x x x x x f . Thus f is decreasing on [1,∞) and so )1()(+>n f n f . Therefore,}{n u is decreasingWe may also show directly that n n u u <+1, that is11)1(122+<+++n nn n This inequality it equivalent to the one we get by cross multiplication:nn n n n n n n n n n n n n n n +<⇔++<+++⇔++<++⇔+<+++2232322221221]1)1[()1)(1(11)1(1Since 1≥n , we know that the inequality 12>+n n is true. Therefore, n n u u <+1and }{n u is decreasing. Condition (2) is readily verified:011lim 1lim lim 2=+=+=∞→∞→∞→nn n n nu n n n n , thus the given series is convergent by the Alternating series Test.1.3.2 Absolute and conditional convergenceIn this section we consider series that have both positive and negative terms. Absolute and conditional convergence. Definition 1.3.1 suppose that the series ∑∞=1k kais not series withpositive terms, if the series∑∞=1k kaformed with the absolutevalue of the terms n a converges, the series ∑∞=1k kais calledabsolutely convergent. The series ∑∞=1k kais called conditionallyconvergent, if the series∑∞=1k kaconverges but∑∞=1k kadiverges.Theorem 1.3.2 if∑kaconverges, the ∑k a converges.Proof for each k, k k k a a a ≤≤-, and therefore k k k a a a 20≤+≤.if∑kaconverges, then∑∑=k ka a22converges, and therefore,by theorem 1.2.3 (the ordinary comparison theorem),∑+)(k ka aconverges. Since k k k k a a a a -+=)(by the theorem1.1.2 (1), we can conclude that∑kais convergence.The above theorem we just proved says that Absolutely convergent series are convergent.As well show presently, the converse is false. There are convergent series that are not absolutely convergent; such series are called conditionally convergent.EX.1.3.4 Prove the following series is absolutely convergent (5141312112)222++-+-Proof If we replace term by it ’s absolute value, we obtain the series (4)131211222++++This is a P series with P=2. It is therefore convergent. This means that the initial series is absolutely convergent.EX.1.3.5 proves that the following series is absolutely convergent: (2)12121212121212118765432+--+--+--Proof if we replace each term by its absolute value, we obtain。

JOURNAL OF FORMALIZED MATHEMATICSVolume1,Released1989,Published2003Inst.of Computer Science,Univ.of BiałystokConvergent Sequences and the Limit of SequencesJarosław KotowiczWarsaw UniversityBiałystokSummary.The article contains definitions and same basic properties of bounded se-quences(above and below),convergent sequences and the limit of sequences.In the articlethere are some properties of real numbers useful in the other theorems of this article.MML Identifier:SEQ_2.WWW:/JFM/Vol1/seq_2.htmlThe articles[1],[6],[3],[5],[7],[2],and[4]provide the notation and terminology for this paper.For simplicity,we adopt the following rules:n,m are natural numbers,r,r1,p,g1,g are real numbers,s1,s′1,s2are sequences of real numbers,y is a set,and f is a real-yielding function.We now state several propositions:(3)1If0<g,then0<g.4(4)If0<g,then g.p(7)If0≤g and0≤r and g<g1and r<r1,then g·r<g1·r1.(9)3−g<r and r<g iff|r|<g.(10)If0<r1and r1<r and0<g,then g.r1(11)If g=0and r=0,then|g−1−r−1|=|g−r|1The propositions(1)and(2)have been removed.2The proposition(5)has been removed.3The proposition(8)has been removed.1c Association of Mizar Users(Def.4)There exists r such that for every n holds r<s1(n).Let us consider f.We say that f is bounded if and only if:(Def.5)f is upper bounded and lower bounded.Let us mention that every real-yielding function which is bounded is also upper bounded and lower bounded and every real-yielding function which is upper bounded and lower bounded is also bounded.The following two propositions are true:(15)4s1is bounded iff there exists r such that0<r and for every n holds|s1(n)|<r.(16)For every n there exists r such that0<r and for every m such that m≤n holds|s1(m)|<r.Let us consider s1.We say that s1is convergent if and only if:(Def.6)There exists g such that for every p such that0<p there exists n such that for every m such that n≤m holds|s1(m)−g|<p.Let us consider s1.Let us assume that s1is convergent.The functor lim s1yields a real number and is defined as follows:(Def.7)For every p such that0<p there exists n such that for every m such that n≤m holds |s1(m)−lim s1|<p.Let us consider s1.Then lim s1is a real number.Next we state a number of propositions:(19)5If s1is convergent and s′1is convergent,then s1+s′1is convergent.(20)If s1is convergent and s′1is convergent,then lim(s1+s′1)=lim s1+lim s′1.(21)If s1is convergent,then r s1is convergent.(22)If s1is convergent,then lim(r s1)=r·lim s1.(23)If s1is convergent,then−s1is convergent.(24)If s1is convergent,then lim(−s1)=−lim s1.(25)If s1is convergent and s′1is convergent,then s1−s′1is convergent.(26)If s1is convergent and s′1is convergent,then lim(s1−s′1)=lim s1−lim s′1.(27)If s1is convergent,then s1is bounded.(28)If s1is convergent and s′1is convergent,then s1s′1is convergent.(29)If s1is convergent and s′1is convergent,then lim(s1s′1)=lim s1·lim s′1.(30)If s1is convergent,then if lim s1=0,then there exists n such that for every m such thatn≤m holds|lim s1|4The propositions(12)–(14)have been removed.5The propositions(17)and(18)have been removed.(34)If s1is convergent and s′1is convergent and for every n holds s1(n)≤s2(n)and s2(n)≤s′1(n)and lim s1=lim s′1,then lim s2=lim s1.(35)If s1is convergent and lim s1=0and s1is non-zero,then s1−1is convergent.(36)If s1is convergent and lim s1=0and s1is non-zero,then lim(s1−1)=(lim s1)−1.(37)If s′1is convergent and s1is convergent and lim s1=0and s1is non-zero,then s′1/s1isconvergent.(38)If s′1is convergent and s1is convergent and lim s1=0and s1is non-zero,then lim(s′1/s1)=lim s′1。

A dedicated effort to determine the complete nucleotide se-quence of the haploid genome in a variety of organisms has been underway since 1990. With this sequence information in hand, geneticists can consult the universal dictionary equating nu-cleotide sequence with amino acid sequence to decide what parts of a genome are likely to be genes. They can also identify genes through matches with nucleotide sequences already known to en-code proteins in other organisms. As a result, they can predict the total number of genes in an organism from the complete nu-cleotide sequence of its genome, and by extension, identify the number and amino acid sequences of all the polypeptides that de-termine phenotype. Knowledge of DNA sequence thus opens up powerful new possibilities for understanding an organism’s growth and development at the molecular level.Studies of the tiny nematode Caenorhabditis elegans illustrate the kind of insights researchers can gain from this DNA-sequence-based approach. C. elegans is a roundworm 1 mm in length that lives in soils throughout the world (Fig. 7.1a). Feeding on bacteria,it grows from fertilized egg to adult—either hermaphrodite or male—in just three days. At the end of this time, each hermaph-rodite produces between 250 and 1000 progeny. Because of its small size, short life cycle, and capacity for prolific reproduction, C. elegans is an ideal subject for genetic analysis.The haploid genome of C.elegans contains110million base pairs distributed among six chromosomes(Fig.7.1b).In the mid-1990s,a group of investigators reported the sequencing and pre-liminary analysis of2.2million base pairs on chromosome ing their knowledge of the concepts explored in this chapter,they found that this2%of the nematode genome carries about480 genes.Interestingly,at least20%of the genes recognized as hav-ing a known function encode molecules that play some role in gene expression: the process by which cells convert DNA se-quence information to RNA and then decode the RNA information to the amino acid sequence of a polypeptide(Fig.7.2).The fact that20%of the genes in this sequenced region encode compo-nents of gene expression suggests the importance of the processCHAPTER7Gene Expression: The Flowof Genetic Information from DNA via RNA to ProteinThe ability of an aminoacyl-tRNA synthetase (red) to couple a particular tRNA (blue)to its corresponding amino acid is central to the molecular machinery that convertsthe language of nucleic acids into the language of proteins.The Genetic Code: How Precise Groupings of the 4 Nucleotides Specify 20 Amino Acids223Figure 7.1 C. elegans:An ideal subject for genetic analysis.(a)Micrograph of several adult worms. (b)Six chromosomes form the haploid genome of C. elegans.The highlighted region depicts a 2.2million base pair portion of chromosome III that has been analyzed and found to encode about 480 genes.(a)T ranslationPolypeptideFigure 7.2Gene expression: The flow of geneticinformation from DNA via RNA to protein.Transcription and translation convert the information encoded in DNA into the order of amino acids in a polypeptide. In transcription, an enzyme known as RNA polymerase catalyzes production of an RNA transcript. In translation, the cellular machinery uses instructions in mRNA to synthesize a polypeptide, following the rules of the genetic code.to the life of the organism.If this ratio holds for the rest of the worm’s genome,about 3600of its estimated 18,000genes gener-ate the machinery that enables genes to be interpreted as proteins.In this chapter, we describe the cellular mechanisms that carry out gene expression. As intricate as some of the details may ap-pear, the general scheme of gene expression is elegant and straightforward: Within each cell, genetic information flows from DNA to RNA to protein.In 1957 Francis Crick proposed that genetic information flows in only one direction, and named his concept of a one-way molecular flow the “Central Dogma” of molecular biol-ogy. As Crick explained, “once ‘information’ has passed into pro-tein, it cannot get out again.”Inside most cells, as the Central Dogma suggests, genetic in-formation flows from one class of molecule to another in two dis-tinct stages (see Fig. 7.2). If you think of genes as instructions written in the language of nucleic acids, the cellular machinery first transcribes a set of instructions written in the DNA dialect to the same instructions written in the RNA dialect. The conversion ofDNA-encoded information to its RNA-encoded equivalent is known as transcription . The product of transcription is a tran-script : a molecule of messenger RNA (mRNA) in prokaryotes;a molecule of RNA that undergoes processing to become an mRNA in eukaryotes. In the second stage of gene expression, the cellular machinery translates the mRNA to its polypeptide equiva-lent in the language of amino acids. This decoding of nucleotide information to a sequence of amino acids is known as transla-tion . It takes place on molecular workbenches called ribosomes,which are composed of proteins and ribosomal RNAs (rRNAs); and it depends on the universal dictionary known as the genetic code , which defines each amino acid in terms of a specific se-quence of three nucleotides. It also depends on transfer RNAs (tRNAs), small RNA adaptor molecules that place specific amino acids at the correct position in a growing polypeptide chain. The tRNAs can bring amino acids to the right place on the translational machinery because tRNAs and mRNAs have complementary nu-cleotides that can base pair with each other.Four general themes emerge from our discussion of gene ex-pression. First, the pairing of complementary bases figures promi-nently in the precise transfer of information from DNA to RNA and from RNA to polypeptide. Second, the polarities of DNA, RNA, and protein molecules help guide the mechanisms of gene expression:the 3′-to-5′transcription of a template DNA strand yields a polar mRNA that grows from its 5′to its 3′end; the 5′-to-3′translation of this mRNA yields a polar protein running from amino terminal to carboxyl terminal. Third, like DNA replication and recombina-tion (discussed in Chapter 5), gene expression requires an input of energy and the participation of several specific proteins at differ-ent points in the process. Fourth, since the accurate one-way flow of genetic information determines protein structure, mutations224CHAPTER 7GENE EXPRESSION: THE FLOW OF GENETIC INFORMATION FROM DNA VIA RNA TO PROTEINthat change this information or obstruct its flow can have dramatic effects on phenotype.As we examine how cells use the sequence information con-tained in DNA to construct proteins, we presentI The genetic code: How triplets of the 4 nucleotidesunambiguously specify 20 amino acids, making it possibleto translate information from a nucleotide chain to asequence of amino acids.I Transcription: How RNA polymerase, guided by basepairing, synthesizes a single-stranded mRNA copy of agene’s DNA template.I Translation: How base pairing between mRNA and tRNAsdirects the assembly of a polypeptide on the ribosome.I A comprehensive example of gene expression in C. elegans.I How mutations affect gene information and expression.THE GENETIC CODE:HOW PRECISE GROUPINGS OF THE4NUCLEOTIDES SPECIFY20AMINO ACIDSA code is a system of symbols that equates information in one language with information in another. A useful analogy for the genetic code is the Morse code, which uses dots and dashes or short and long sounds to transmit messages over radio or tele-graph wires. Various groupings of the dot-dash/short-long symbols represent the 26 letters of the English alphabet. Be-cause there are many more letters than the two dot or dash symbols, groups of up to four dots, four dashes, and varying combinations of the two represent some letters. And because anywhere from one to four symbols specify each letter, the Morse code requires a symbol for “pause” to signify where one letter ends and the next begins.In the Genetic Code, a Triplet Codon Represents Each Amino AcidThe language of nucleic acids is written in four nucleotides—A,G,C,and T in the DNA dialect;A,G,C,and U in the RNA dialect—while the language of proteins is written in amino acids.To understand how the sequence of bases in DNA or RNA encodes the order of amino acids in a polypeptide chain,it is essential to know how many distinct amino acids there are.Watson and Crick produced the now accepted list of the20amino acids that are genetically encoded by DNA or RNA sequence over lunch one day at a local pub.They cre-ated the list by analyzing the amino acid sequence of a vari-ety of naturally occurring polypeptides.Amino acids that are present in only a small number of proteins or in only certain tissues or organisms did not qualify as standard building blocks;Crick and Watson correctly assumed that such amino acids arise when proteins undergo modification after their synthesis.By contrast,amino acids that are present in most, though not necessarily all,proteins made the list.The ques-tion then became:How can4nucleotides encode20amino acids?Just as the Morse code conveys information through dif-ferent groupings of dots and dashes, the 4 nucleotides encode 20 amino acids through specific groupings of A, G, C, and T or A, G, C, and U. Researchers initially arrived at the number of letters per grouping by deductive reasoning, and later con-firmed it by experiment. They reasoned that if only one nu-cleotide represented an amino acid, there would be informa-tion for only four amino acids: A would encode one aminoacid; G, a second amino acid; C, a third; and T, a fourth. If twonucleotides represented each amino acid, there would be 42ϭ16 possible combinations of couplets.Of course, if the code consisted of groups containing oneor two nucleotides, it would have 4 ϩ16 ϭ20 groups and could account for all the amino acids, but there would be noth-ing left over for the pause denoting where one group ends and the next begins. Groups of three nucleotides in a row would provide 43ϭ64 different triplet combinations, more than enough to code for all the amino acids. If the code consisted of doublets and triplets, a signal denoting pause would once again be necessary. But a triplets-only code would require no symbol for “pause” if the mechanism for counting to three and distinguishing among successive triplets were very reliable.Although this kind of reasoning—explaining the un-known in terms of the known by looking for the simplest pos-sibility—generates a theory, it does not prove it. As it turnedout, however, the experiments described later did indeeddemonstrate that groups of three nucleotides represent all 20amino acids. Each nucleotide triplet is called a codon.Eachcodon, designated by the bases defining its three nucleotides,specifies one amino acid. For example, GAA is a codon forglutamic acid (Glu), and GUU is a codon for valine (Val). Be-cause the code comes into play only during the translation partof gene expression, that is, during the decoding of messengerRNA to polypeptide, geneticists usually present the code in theRNA dialect of A, G, C, and U, as depicted in Fig. 7.3. How-ever, when speaking of genes, they can substitute T for U toshow the same code in the DNA dialect.If you knew the sequence of nucleotides in a gene or itstranscript as well as the sequence of amino acids in the corre-sponding polypeptide, you could deduce the genetic codewithout understanding how the cellular machinery uses thecode to translate from nucleotides to amino acids. Althoughtechniques for determining both nucleotide and amino-acidsequence are available today, this was not true when re-searchers cracked the genetic code in the 1950s and 1960s. Atthat time, they could establish a polypeptide’s amino-acid se-quence, but not the nucleotide sequence of DNA or RNA. Be-cause of their inability to read nucleotide sequence, they usedAThe Genetic Code: How Precise Groupings of the 4 Nucleotides Specify 20 Amino Acids225Figure 7.3The genetic code: 61 codons represent the 20 amino acids, while 3 codons signify stop.To read the code, find the first letter in the left column, the second letter along the top, and the third letter in the right column; this reading corresponds to the 5′-to-3′direction along the mRNA. Although most amino acids are encoded by two or more codons, the genetic code is unambiguous because each codon specifies only one amino acid.an assortment of genetic and biochemical techniques to fathom the code. They began by examining how different mu-tations in a single gene affected the amino-acid sequence of the gene’s polypeptide product, using the abnormal (specific mutations) to understand the normal (the general relationship between genes and polypeptides).Mapping Studies Confirmed That a Gene’s Nucleotide Sequence Is Colinear with a Polypeptide’s Amino-Acid SequenceWe have seen that DNA is a linear molecule with base pairs following one another down the intertwined chains. Proteins, by contrast, have complicated three-dimensional structures. Even so, if unfolded and stretched out from amino terminus to carboxyl terminus, proteins have a one-dimensional, linear structure—a specific sequence of amino acids. If the informa-tion in a gene and its corresponding protein are colinear, the consecutive order of bases in the DNA from the beginning to the end of the gene would stipulate the consecutive order of amino acids from one end to the other of the outstretched pro-tein. Note that this hypothesized relationship implies that both a gene and its protein product have definite polarities with an invariant relation to each other.Charles Yanofsky, in studying the Escherichia coli gene for a subunit of the enzyme tryptophan synthetase, was the first to compare maps of mutations within a gene to the par-ticular amino-acid substitutions that resulted. He began by generating a large number of trpϪauxotrophic mutants that carried mutations in the trpA gene for the tryptophan syn-thetase subunit. He next made a fine structure recombinational map of these mutations; and then he purified and determined the amino acid sequence of the mutant tryptophan synthetase subunits. As Fig. 7.4a illustrates, Yanofsky’s data showed that the order of mutations mapped within the DNA of the gene by recombination was colinear with the positions of the amino-acid substitutions occurring in the resulting mutant proteins. Genetic Analysis Revealed That Nonoverlapping Codons Are Set in a Reading FrameBy carefully examining the results of his analysis,Yanofsky, in addition to confirming the existence of colinearity,de-duced key features of codons and helped establish many pa-rameters of the genetic code relating nucleotides to amino acids.A Codon Is Composed of More Than One Nucleotide Yanofsky observed that different point mutations(changes in only one nucleotide pair)may affect the same amino acid.In one example shown in Fig.7.4a,mutation#23changed the glycine(Gly)at position211of the wildtype polypeptide chain to arginine(Arg),while mutation#46yielded glutamic acid(Glu) at the same position.In another example,mutation #78changed the glycine at position234to cysteine(Cys), while mutation#58produced aspartic acid(Asp)at the same position.In both cases,Yanofsky also found that recombina-tion could occasionally occur between two mutations that changed the identity of the same amino acid,and such re-combination would produce a wildtype tryptophan syn-thetase gene(Fig.7.4b).Because the smallest unit of recombination is the base pair,two mutations capable of re-combination—in this case,in the same codon because they affect the same amino acid—must be in different(although nearby)nucleotides.Thus,a codon contains more than one nucleotide.Each Nucleotide Is Part of Only a Single CodonAs Fig. 7.4a illustrates, each of the point mutations in the tryp-tophan synthetase gene characterized by Yanofsky alters the identity of only a single amino acid. This is also true of the point mutations examined in many other genes, such as the hu-man genes for rhodopsin and hemoglobin (see Chapter 6). Since point mutations change only a single nucleotide pair and most point mutations affect only a single amino acid in a polypeptide, each nucleotide in a gene must influence the identity of only a single amino acid. If, on the contrary, a nu-cleotide were part of more than one codon, a mutation in that nucleotide would affect more than one amino acid.226CHAPTER 7GENE EXPRESSION: THE FLOW OF GENETIC INFORMATION FROM DNA VIA RNA TO PROTEIN1 m.u.N C2114926822151234Figure 7.4Experiments analyzing the E. coli gene for a subunit of tryptophan synthetase confirm colinearity and reveal significant features of the genetic code.(a)A genetic map of the trpA gene of E. coli,identifying the amino-acid substitutions that characterize several of Yanofsky’s mutant strains. The positions of the mutations and amino-acid substitutions are colinear. These mutations change only a single amino acid, suggesting that each nucleotide is part of only a single codon. (b)Confirmation that codons must include two or more base pairs came from crosses between two strains that carried an altered amino acid at the same position. Since wildtype progeny occasionally appeared, each strain had a point mutation at a slightly different site. Crossing-over between the mutant sites could produce a wildtype allele.A Codon Is Composed of Three Nucleotides,and the Designated Starting Point for Each Gene Establishes the Reading Frame for These Triplets Although the most efficient code that would allow 4 nu-cleotides to specify 20 amino acids requires 3 nucleotides per codon, more complicated scenarios are possible. Francis Crick and Sydney Brenner obtained convincing evidence for the triplet nature of the genetic code in studies of mutations in the bacteriophage T4 rIIB gene (Fig. 7.5). They induced the mu-tations with proflavin, an intercalating mutagen that can insert itself between the paired bases stacked in the center of the DNA molecule (Fig. 7.5a). Their assumption was that proflavin would act like other mutagens, causing single-base substitutions. If this were true, it would be possible to gener-ate revertants through treatment with any mutagen. Surpris-ingly, genes with proflavin-induced mutations did not revert to wildtype upon treatment with other mutagens known to cause nucleotide substitutions. Only further exposure to proflavin caused proflavin-induced mutations to revert to wildtype (Fig.7.5b). Crick and Brenner had to explain this observation be-fore they could proceed with their phage experiments. With keen insight, they correctly guessed that proflavin does not cause base substitutions; instead, it causes insertions or dele-tions. This hypothesis explained why base-substituting muta-gens could not revert proflavin-induced mutations; it was also consistent with the structure of proflavin. By intercalating be-tween base pairs, proflavin would distort the double helix and thus interfere with the action of enzymes that function in the repair, replication, or recombination of DNA, eventually caus-ing the deletion or addition of one or more nucleotide pairs to the DNA molecule.Crick and Brenner began their experiments with a partic-ular proflavin-induced rIIBϪmutation. They next treated this mutant strain with more proflavin to isolate an rIIBϩrevertant (see Fig. 7.5b), and showed that the revertant’s chromosome actually contained two different rIIBϪmutations: One was the original mutation (FC0 in the figure); the other was newly in-duced (FC7). Either mutation by itself yields a mutant pheno-type, but their simultaneous occurrence in the same gene yielded an rIIBϩphenotype. Crick and Brenner reasoned that if the first mutation was the deletion of a single base pair, rep-resented by the symbol (Ϫ), then the counteracting mutation must be the insertion of a base pair, represented as (ϩ). The restoration of gene function by one mutation canceling an-other in the same gene is known as intragenic suppression. On the basis of this reasoning, they went on to establish T4 strains with different numbers of (ϩ) and (Ϫ) mutations in the same chromosome. Figure 7.5c tabulates the phenotypes asso-ciated with each combination of proflavin-induced mutations.I n analyzing the data, Crick and Brenner assumed that each codon is a trio of nucleotides, and for each gene there is a single starting point. This starting point establishes a read-Three single base insertions ( + + + )Single base insertion (+)Single base deletion (–)ATG AAC AA GCG C G G G GAA GCG GACATG AAC AA T GCG C C G G A G GAA GCG GAC ATG AAC AAT GCG CCG GAG GAA GCG GAC ATG AAC AAT G G CGC T CGG C AG GAA GCG GACATG AAC AAT GCG CCG GAG GAA GCG GAC ATG AAC AAT G G CGCCG GAG GAA GCG GACATG AAC AA T GCG CCG GAG GAA GCG GAC ATG AAC AA GCG CCG GAG GAA GCG GACG GT CFigure 7.5Studies of frameshift mutations in thebacteriophage T4 rIIB gene show that codons consist of (e)228CHAPTER 7GENE EXPRESSION: THE FLOW OF GENETIC INFORMATION FROM DNA VIA RNA TO PROTEINing frame: the partitioning of groups of three nucleotides such that the sequential interpretation of each succeeding tripletgenerates the correct order of amino acids in the resultingpolypeptide chain. If codons are read in order from a fixedstarting point, one mutation will counteract another if the twoare equivalent mutations of opposite signs; in such a case, eachinsertion compensates for each deletion, and this counterbal-ancing restores the reading frame. The gene would only regainits wildtype activity, however, if the portion of the polypeptideencoded between the two mutations of opposite sign is not re-quired for protein function, because in the double mutant, thisregion would have an improper amino-acid sequence. Simi-larly, if a gene sustains three or multiples of three changes ofthe same sign, the encoded polypeptide can still function, be-cause the mutations do not alter the reading frame for the ma-jority of amino acids (Fig. 7.5d). The resulting polypeptidewill, however, have one extra or one fewer amino acid thannormal (designated by three plus signs or three minus signs,respectively), and the region encoded by the part of the genebetween the first and the last mutations will not contain thecorrect amino acids.By contrast, a single nucleotide inserted into or deletedfrom a gene alters the reading frame and thereby affects theidentity of not only one amino acid, but of all other aminoacids beyond the point of alteration (Fig. 7.5e). Changes thatalter the grouping of nucleotides into codons are calledframe shift mutations: they shift the reading frame for all codons beyond the point of insertion or deletion, almost al-ways abolishing the function of the polypeptide product.A review of the evidence tabulated in Fig. 7.5c supportsall these points. A single (Ϫ) or a single (ϩ) mutation de-stroyed the function of the rIIB gene and produced an rIIBϪphage. Similarly, any gene with two base changes of the samesign (ϪϪor ϩϩ), or with four or five insertions or deletionsof the same sign (for example, ϩϩϩϩ) also generated a mu-tant phenotype. However, genes containing three or multiplesof three mutations of the same sign (for example, ϩϩϩor ϪϪϪϪϪϪ) as well as genes containing a (ϩϪ) pair of mutations generated rIIBϩwildtype individuals. In these lastexamples, intragenic suppression allowed restitution of thereading frame and thereby restored the lost or aberrant geneticfunction produced by other frameshift mutations in the gene. Most Amino Acids are Specified by More Than One Codon As Fig.7.5illustrates,intragenic suppression occurs only if in the region between two frameshift mutations of opposite sign, a gene still dictates the appearance of amino acids,even if these amino acids are not the same as those appearing in the normal protein.If the frameshifted part of the gene encodes instructions to stop protein synthesis,for example,by introducing a triplet of nucleotides that does not correspond to any amino acid,then wildtype polypeptide production would not continue.This is because polypeptide synthesis would stop before the compen-sating mutation could reestablish the correct reading frame.The fact that intragenic suppression occurs as often as itdoes suggests that the code includes more than one codon for some amino acids. Recall that there are 20 common amino acids but 43ϭ64 different combinations of three nucleotides. If each amino acid corresponded to only a single codon, there would be 64 Ϫ20 ϭ44 possible triplets not encoding an amino acid. These noncoding triplets would act as “stop” sig-nals and prevent further polypeptide synthesis. I f this hap-pened, more than half of all frameshift mutations (44/64) would cause protein synthesis to stop at the first codon after the mutation, and the chances of extending the protein each amino acid farther down the chain would diminish exponen-tially. As a result, intragenic suppression would rarely occur. However, we have seen that many frameshift mutations of one sign can be offset by mutations of the other sign. The distances between these mutations, estimated by recombination fre-quencies, are in some cases large enough to code for more than 50 amino acids, which would be possible only if most of the 64 possible triplet codons specified amino acids. Thus, the data of Crick and Brenner provide strong support for the idea that the genetic code is degenerate:Two or more nucleotide triplets specify most of the 20 amino acids (see Fig. 7.3). Cracking the Code: Biochemical Manipulations Revealed Which Codons Represent Which Amino AcidsAlthough the genetic experiments just described enabled re-markably prescient insights about the nature of the genetic code,they did not make it possible to assign particular codons to their corresponding amino acids.This awaited the discovery of messenger RNA and the development of techniques for syn-thesizing simple messenger RNA molecules that researchers could use to manufacture simple proteins in the test tube. The Discovery of Messenger RNAs, Moleculesfor Transporting Genetic InformationIn the1950s researchers exposed eukaryotic cells to amino acids tagged with radioactivity and observed that protein syn-thesis incorporating the radioactive amino acids into polypep-tides takes place in the cytoplasm,even though the genes for those polypeptides are sequestered in the cell nucleus.From this discovery,they deduced the existence of an intermediate molecule,made in the nucleus and capable of transporting DNA sequence information to the cytoplasm where it can di-rect protein synthesis.RNA was a prime candidate for this in-termediary information-carrying molecule.Because of RNA’s potential for base pairing with a strand of DNA,one could imagine the cellular machinery copying a strand of DNA into a complementary strand of RNA in a manner analogous to the DNA-to-DNA copying of DNA replication.Subsequent stud-ies in eukaryotes on the incorporation of radioactive uracil(a base found only in RNA)into molecules of RNA showed that although the molecules are synthesized in the nucleus,at least some of them migrate to the cytoplasm.Among those RNA molecules that migrate to the cytoplasm are the messenger RNAs,or mRNAs,depicted in Fig.7.2.They arise in the nu-cleus from the transcription of DNA sequence information through base pairing and then move,after processing,to the。

APPENDIX ALimit of Sequences of Numberssequence limit , . limit , .Let(a n)be a sequence of real numbers.Definition A.1.(1)Suppose that there is a real number U satisfying the follow-ing two conditions:(a)For allε>0,there exists an N>0such that for all n≥N,a n<U+ε.(b)Givenε>0and given m>0,there exists an n>m such thata n>U−ε.Then U is called the limit superior(or upper limit)of(a n)and we writeU=lim supn→∞a n=limn→∞a n.(2)If(a n)is not bounded above,we definelim supn→∞a n=+∞.(3)If(a n)is bounded above,but not bounded below and if(a n)has nofinite limitsuperior,then we saylim supn→∞a n=−∞.173174 A.LIMIT OF SEQUENCES OF NUMBERS(4)The limit inferior(or lower limit)of(a n)is defined bylim inf n→∞a n=limn→∞a n=−lim supn→∞(−a n)., . Remark .Remark A.2.(a)in Definition A.1(1)=⇒all terms after A N lie to the left of U+ε;(b)in Definition A.1(1)=⇒infinite many terms lie to the right of U−ε.Remark A.3.Sincelim sup n→∞a n=limn→∞supm≥na m=infnsupm≥na m,lim inf n→∞a n=limn→∞infm≥na m=supninfm≥na m,the limit superior and limit inferior are unique in R∪{+∞,−∞}.Example A.4.(1)a n=(−1)n.Thenlim sup a n=1,lim inf a n=−1.(2)a n=(−1)n1+1n.Thenlim sup a n=1,lim inf a n=−1.(3)a n=(−1)n·n.Thenlim sup a n=∞,lim inf a n=−∞.(4)a n=n2sin212nπ.Thenlim sup a n=∞,lim inf a n=0.Theorem A.5.(1)lim infn→∞a n≤lim supn→∞a n.A.LIMIT OF SEQUENCES OF NUMBERS175(2)The sequence(a n)converges if and only iflim sup n→∞a n=lim infn→∞a n<∞.Moreover,lim sup n→∞a n=lim infn→∞a n=limn→∞a n.(3)The sequence(a n)diverges to+∞if and only iflim sup n→∞a n=lim infn→∞a n=+∞.(4)The sequence(a n)diverges to−∞if and only iflim sup n→∞a n=lim infn→∞a n=−∞.lim sup lim inf . Definition A.1 , , lim sup lim inf . Remark A.3 , , , .。