微积分(下)英文教材

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.Whil e 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 the sum ofthe 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 theform 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 that n 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 x x 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 this instance, wehave -∞=--=∞→∞→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 kSolution 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 series converges to 1.1)211(lim lim =+-=∞→∞→n S n n n therefore 1)2)(1(1=++∑∞=n k kEX.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 the sequence }{n S ofpartial sums exists. Namelyl a S nk k n n n ==∑=∞→∞→0lim limObviously.lim lim 01l a S nk k n n n ==∑=∞→-∞→since1--=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 for which 0lim =∞→k k a . Forexample, the series.....1. (2)11111++++=∑∞=nkk . Since it issequence }{n S of partial sum nnn nS 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 series diverges. EX.1.1.4 Determine whether or not the series ∑∞=021k kSolution1the given series is a geometric series.121,)21(00<==∑∑∞=∞=x and xk k k k, by proposition 1.1.1 we know that seriesconverges.Solution 2 ,21 (4)12111-++++=n n S ①,2121.........21212121132n n n S +++++=-② ①-② (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 two series.(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 that n n n c b a ≤≤ for,k n ≥(k is a fixed integer), then }{n b also converges to l .Ex.1.1.6 show that 0sin lim3=∞→nnn . Solution For ,1≥n ,1)sin (13n n 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 series1nn 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. Derivethe following results from the geometricseries 221(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 upper bound. 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 numbers0≥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 to determine when a series with positive terms converges.The convergence or divergence of a series with nonnegative terms is usually deduced by comparison with a series of known behavior.S 1 S 2 S n STheorem 1.2.3(The Ordinary Comparison Test) Let1nn a∞=∑and1nn b∞=∑be two series, with 0n a ≥ for all n and 0n b ≥ for all n. Assume thatthere is a numbers 0c >, such that n n a cb ≤ for all n, and that 1nn 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 partial sums 12n a a a +++….The least upper bound of these sums is therefore 1n n c b ∞=≤∑, thus proving ourtheorem.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∞=∑ betwo series, with n a and 0n b ≥ for all n. Assume that there is a number0c > such that n n a cb ≥ for all n sufficiently large, and1nn b∞=∑ does notconverge, then1nn a∞=∑ diverges.Proof. Assume n n a cb ≥for 0n n ≥, since 1nn b∞=∑diverges, we canmake the partial sum0001Nnn n N n n bb b b +==+++∑…arbitrarily large as N becomes arbitrarily large. But 0NNNn n n n n n n n n a cb c b ===≥=∑∑∑.Hence the partial sum121NnN n aa a a ==+++∑… are arbitrarily large as Nbecomes arbitrarily large, are hence1nn a∞=∑ diverges, as was to be shown.Remark on notation you have easily seen that for each 0j ≥, 0kk a∞=∑converges iff1kk j a∞=+∑ converges. This tells us that, in determining whetheror 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.For instance, 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 example122k k ∞==∑, 1112k k ∞==∑, 21122k k ∞==∑, 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. Our partial 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 partialsum 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 series converges.Generally we have the following result: The series 1111111234p p p p pn nn ∞==++++++∑……, where p is a constant, is called a p-series.Proposition1.2.1. If 1p >, the p-series converges; and if 1p ≤, 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 that23111122n n n n≥=+. Since 11n n ∞=∑ does not converge, it follows that the series 2311n n n ∞=+∑ does not converge either. Namely this series 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 our series with 21n ∑ tosee converges, because∑21n converges and 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∑ beseries with positive terms. If lim()k k ka lb →∞=, where l is some positivenumber, thenka∑ andkb∑converge or diverge together.Proof Choose ε between 0 and l , since k ka lb →, we know for allk sufficiently large (for all k greater than some 0k ) ||kka lb ε-<. For such k we have kka l lb εε-<<+, and thus ()()k k k l b a l b εε-<<+ this last inequality is what we needed. (1) Ifka∑converges, then()kl b ε-∑converges, and thuskb∑converges.(2) Ifkb∑converges, then()kl bε+∑converges, and thuska∑converges.To apply the limit comparison theorem to a series k a ∑, we must first find a seriesk b ∑of known behavior for whichkka b converges to a positive number.Ex 1.2.6 Determine whether the series sinkπ∑converges ordiverges.Solution Recall that as sin 0,1x x x →→. As ,0k kπ→∞→ and thus sin 1k kππ→. Sincek π∑diverges, sosin()k π∑diverges.Ex 1.2.7 Determine whether the series 2529k k k k +converges ordiverges.Solution For large value of k , 5k dominates the numerator and2k k the denominator, thus, for such k ,2529k k k k+differs 225522k kk k =. Since 2222222015100510200192291004512k k k kk k k k k k k k k +++÷==→+++And2255122k k =∑∑converges, this series converges.Theorem 1.2.6 Let ka∑ andkb∑ be series with positive termsand suppose thus0kka b →, then (1) If kb∑converges, then ka∑converges.(2) If ka ∑diverges, thenkb ∑diverges.(3) If ka∑converges, then kb∑may converge or diverge.(4) Ifkb∑diverges, thenka∑may converge or diverge.[Parts (3) and (4) explain why we stipulated 0l >in theorem 1.2.5]1.2.2 The root test and the ratio test Theorem 1.2.7 (the root test, Cauchy test) let ∑kabe a series withnonnegative terms and suppose thatρ==∞→∞→k kk k k k a a 1lim lim , ifρ<1,∑kaconverges, if ρ>1,∑kadiverges, if ρ=1, the test is inconclusive.Proof we suppose first ρ<1 and choose μ so that 1<<u ρ. Sinceρ→kk a 1)(, we have μ<k ka 1, for all k sufficiently large thus k k a μ< for allk sufficiently large since∑kμconverges (a geometric series with0<1<μ), we know by theorem 1.2.5 that∑kaconverges.We suppose now that 1>ρand choose μso that 1>>u ρ. sinceρ→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 ) the theorem1.2.6 tell us that∑kadiverges.To see the inconclusiveness of the root test when 1=ρ, note that1)(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 ∑kk )(ln 1converges ordiverges.Solution For the series ∑kk )(ln 1, applying the root test we have 0ln 1lim)(lim 1==∞→∞→ka k kk k , the series converges. EX.1.2.8 Determine whether series ∑3)(2k kconverges or diverges.Solution For the series ∑k k)3(2, applying the root test, we have1212]1[2)1(.2)(3331>=⨯→==k k kk k k a . So the series diverges.EX1.2.9 Determines whether the series kk∑-)11(converges ordiverges.Solution in the case of kk ∑-)11(, we have 111)(1→-=ka k k . Ifapplying the root test, it is inconclusive. But since k k ka )11(-=converges toe1and 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 ∑kabe a serieswith 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 byinduction ,N k k N a C a ≤+Thusca c c c c a a c a c ca a aNk N N k 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 .1lim1<=+∞→λnn n a a EX.1.2.10 show that the series∑∞=13n n nconverges. 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 k k 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 lim1Since 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 diverge together.3. Let nn n a a 1lim+∞→=ρ. The ratio Test says that a series ∑kaof positive termsconverges 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 k 9.∑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 thatna a n n 111-≥+ for all n. show that the series ∑nadiverges.1.3 Alternating series, Absolute convergence and conditional convergenceIn this section we consider series that have both positive and negative terms.1.3.1 Alternating series and the tests for convergenceThe 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 find 3s we add 3u , so 3s is to the right of 2s . But, since 3u <2u , 3s is to the 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 are converging 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. We first 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 sBut 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 theor em. Let’s call it is limit s, that is, s s n n =∞→2lim Now we compute the limit of the odd partialsums: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 and divergence.Solution the given series is alternating but043143lim 143lim lim≠=-=-=∞→∞→∞→nn n u n n n n So condition (2) is not satisfied. Instead, we look at the limit of the nthterm 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 or divergence.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 function 1)(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 n n nThis 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 with positiveterms, if the series∑∞=1k kaformed with the absolute value of the termsn a converges, the series∑∞=1k kais called absolutely convergent. The series∑∞=1k kais called conditionally convergent, if the series∑∞=1k kaconvergesbut∑∞=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, bytheorem 1.2.3 (the ordinary comparison theorem),∑+)(k ka aconverges.Since k k k k a a a a -+=)(by the theorem 1.1.2 (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 convergentseries 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 the series: (2)12121212121212118765432+++++++=+This is a convergent geometric series. The initial series is therefore absolutely convergent.Ex.1.3.6 proves that the following series is only conditionally convergent:∑∞=-=++-+-+-1)1(.............61514131211n nnProof the given series is convergent. Since (1) ,1111nu n u n n =<+=+(2) 01lim lim ==∞→∞→n u n n n , So this series is convergent by the alternating series test, but it is not absolutely.Convergent: if we replace each term by it is absolute value, we obtain the divergent harmonic series:。