微积分(下)英文教案资料

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。