Contents. 7 Sequences and Series. 7.1 Sequences and Convergence. Calculus II (part 3): Sequences and Series (by Evan Dummit, 2015, v. 2.

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1 Calculus II (part 3): Sequeces ad Series (by Eva Dummit, 05, v..00) Cotets 7 Sequeces ad Series 7. Sequeces ad Covergece Iite Series Geometric Series Telescopig Series Positive Series: Itegral Test, Compariso Tests, Ratio ad Root Tests Itegral Test Compariso Test Limit Compariso Test Ratio Test Root Test Geeral Series: Alteratig Series Test, Absolute ad Coditioal Covergece Alteratig Series Test Absolute ad Coditioal Covergece Rearragemets of Iite Series Further Examples of Series Covergece Tests Sequeces ad Series Our rst goal i this chapter is to itroduce the otio of a coverget sequece ad to discuss the closely related cocept of a series. We give a deitio for iite series, ad the to discuss how to calculate the value of iite sums i some cases. We the tur our attetio to the questio of whether a give iite series coverges. We discuss the most commoly used covergece tests for positive series (the Itegral Test, Compariso Tests, Ratio Test, ad Root Test) ad the discuss geeral series whose terms may be positive or egative (i particular, alteratig series) ad the ideas of absolute ad coditioal covergece. We will ot dwell extesively o the techical details ivolved i the justicatios of all of the results: although we will give proofs of the mai results whe feasible, learig the details of the proofs is far less importat tha uderstadig how the results are used. Our goal is primarily to state the tests, illumiate the uderlyig ideas behid them, ad explai how they are used. To this ed, we close our discussio with a extesive array of examples illustratig the covergece tests. 7. Sequeces ad Covergece Deitio: A sequece is a (ite or iite) ordered list of umbers a, a, a 3, a 4,..., a,... We will usually be cocered with iite sequeces. The sequece,, 3, 4,... is the sequece of positive itegers, whose th term is deed as a =. The sequece,, 3, 4,... is the sequece of reciprocals of positive itegers, whose th term is a =. Aother sequece is,,,,,,..., whose th term is a = ( ) +.

2 Oe of the most immediate questios for a iite sequece is how does this sequece behave as we go further ad further out? I other words, what happes to a as? Oe possibility is that the terms could approach (closer ad closer) to some xed limitig value L. Notice, for example, that the terms i the sequece,, 3,,... are gettig closer ad closer to 0. 4 Aother possibility is that the terms could grow without boud, like the terms i the sequece,, 3, 4,... Yet aother possibility is that the terms could just bouce aroud ad ot settle dow to aythig, like the terms i the sequece,,,,,,... Ultimately, we are askig about whether this sequece coverges to a limit. This is very much the same questio we could ask about a fuctio f(x): amely, what happes to the value of f(x) as x? Deitio: We say a sequece a, a,... coverges to the limit L if, for ay ɛ > 0 (o matter how small) there exists some positive iteger N such that for every N, it is true that a L < ɛ. If there is o value of L such that the sequece a coverges to L, the we say the sequece a diverges. Ituitively, the deitio says that the sequece coverges to L if the terms of the sequece evetually get ad stay arbitrarily close to L. This deitio is almost idetical to the formal ɛ δ deitio of the limit of a fuctio. Like with that deitio, we use it primarily as a startig poit. Example: The sequece,, 3, 4, 5,... whose th term is a = coverges to 0. I the formal deitio, we ca take N = /ɛ: the for ay N, we see that a L = N ɛ < ɛ, as desired. Example: The sequece,,,,,... whose th term is a = ( ) forever betwee + ad. diverges, sice it alterates This series, i a imprecise way, really is tryig to coverge to 'two values' (amely, ad ). Remark: The covergece of a sequece is ot aected if we remove a ite umber of terms from the sequece. So, for example, the sequece a, a, a 3, a 4, a 5, has exactly the same covergece properties as the sequece a 5, a 6, a 7,. As with limits of a fuctio, the limit of a sequece obeys a umber of simple rules (sometimes collectively called the Limit Laws). Theorem (Limit Laws for Sequeces): Let {a } ad {b } be sequeces of real umbers ad A, B be real umbers. If lim a = A ad lim b = B, the the followig properties hold: The additio rule: lim [a + b ] = A + B. The subtractio rule: lim [a b ] = A B. The multiplicatio rule: lim [a b ] = A B. Note that the multiplicatio rule yields as a special case (whe b is idetically equal to a costat c) the costat-multiplicatio rule: lim [c a ] = c A, where c is ay real umber. [ ] a The divisio rule: lim = A. provided that B is ot zero. b B The squeeze rule (also called the sadwich rule): If a b c ad lim a = L = lim c (meaig that both limits exist ad are equal to L) the lim b exists ad is also L. The proofs of these results follow i the same way as for limits of fuctios. We omit the details. We also have a few additioal results cocerig limits of sequeces:

3 The Mootoe Covergece Theorem: If the sequece a, a,... is mootoe icreasig ad bouded above, the it coverges. A sequece is mootoe icreasig if a < a < a 3 <, ad a sequece is bouded above if there exists some M with all a i M. This property is almost equivalet to part of the deitio of the real umbers: it follows from what is called the least upper boud property. By multiplyig everythig by, the theorem also says that if a sequece of real umbers is mootoe decreasig ad bouded below, the it has a limit. Example: The sequece a with a = coverges, because the terms are mootoe icreasig, ad they are all bouded above by because the square is always positive. (I fact, the limit of this sequece is.) The Cotiuous Fuctio Theorem: If f(x) is ay cotiuous fuctio ad {a } is ay coverget sequece with lim a = A, the lim f(a ) = f(a). This theorem is ituitively very atural: for a fuctio to be cotiuous, it must be the case that as we approach ay poit x = A, the value of the fuctio must get closer ad closer to f(a). This is precisely the behavior captured by the coverget sequece. Example: The sequece b with b = / coverges to, because if we set a = ad f(x) = x, the b = f(a ). Sice lim a = 0, the Theorem idicates that lim f(a ) = 0 =. Just like with limits of fuctios, we ca sometimes be more precise about the way i which a sequece diverges: if the terms grow very large ad positive, for example, it is atural to wat to say that the sequece diverges to +. We will say that a sequece a, a,... diverges to + if for ay M > 0 (o matter how large) there exists some N such that for every N, it is true that a > M. We will also say that a sequece a, a,... diverges to if for ay M > 0 (o matter how large) there exists some N such that for every N, it is true that a < M. These are just formalisms that give a more careful meaig to the idea of the terms get ad stay arbitrarily large ad positive or the terms get ad stay arbitrarily large ad egative. Frequetly, i computig limits of sequeces, the terms of the sequece are the values f(), f(), f(3),... for some simple fuctio f(x). There is a very atural relatio betwee the limit of the sequece ad the limit of the fuctio: Propositio: If f(x) is ay fuctio such that lim f(x) = L, the the limit of the sequece f(), f(), f(3), x... is equal to L. Also, if the limit of the fuctio is (or ), the so is the limit of the sequece. The idea is simply that if the sequece f(), f(), f(3),... did ot have limit L, the this would cotradict the statemet that lim f(x) = L, sice both otios of covergece are capturig the idea x that the values of f(x) must get ad stay close to L for large x. Example: The sequece,, 3,... whose th term is a = f(), f(3),... with f(x) =, ad lim x x x = 0. Example: The sequece,, 4,... whose th term is a = f(), f(3),... with f(x) = x, ad lim x x =. Frequetly, it is useful to ivoke L'Hôpital's Rule i combiatio with this result. coverges to 0 : it is the sequece f(), diverges to + : it is the sequece f(), Recall that L'Hôpital's Rule says that if f ad g are dieretiable fuctios, ad form 0 0 or f(), the lim g() = lim f (), assumig the secod limit exists. g () f() lim g() is of the

4 The result above says that we ca use L'Hôpital's Rule to compute limits of sequeces, provided they are a quotiet of the ecessary form. Example: The sequece, 3 3, 5 4, 7 5, 9 6,... whose th term is a = coverges to : it is the sequece f(), f(), f(3),... with f(x) = + we used l'hôpital's rule i the equality labeled l'h. +. We the compute lim + l'h = lim =, where x 7. Iite Series If a, a,... is a sequece, recall the summatio otatio terms of the form a, where rus from to k. a = a + a + + a k : it meas add up all We would like to give meaig to a iite sum: somethig like a = a + a + + a k +. As writte, this does't really make sese: how do we add iitely may thigs? The aswer is: we add the rst k of them, ad the cosider what happes as k. Deitio: If a, a, a 3,... is a iite sequece, we dee the value (or sum) of the iite series to be the limit lim k S k, where S k = a is the kth partial sum of the series (amely, the sum of the rst terms i the series). We say the series coverges to L if the limit lim k S k exists ad equals L, ad we say it diverges otherwise. (We also iclude the possibility that it could diverge to or to.) Importat Note: Notice that a series is dieret from a sequece: a series is a sum, while a sequece is a list of umbers. Example: The series coverges to : the sequece of partial sums is, 3 4, 7 8, 5 6, 3 3, 63 64,.... We see that the kth partial sum is S k =, ad sice lim k S k = lim ( ) k k k =, the series coverges to. Example: The series ( ) diverges : the sequece of partial sums is, 0,, 0,, 0,..., which alterates forever ad does ot coverge. Example: The series diverges to + : the sequece of partial sums is, 6, 4, 30, 6,.... We see that the kth partial sum is S k = k+, which diverges to. (Alteratively, if we did't see the patter, clearly the kth term is bigger tha k.) I order for a series to coverge, the terms of the series must evetually be small. Explicitly: Test (Divergece Test): The series a diverges if lim a 0. If the limit exists ad is positive, the the series diverges to + ; if it is egative, the series diverges to. The limit does ot equal zero part icludes the case where the limit does't exist. Ituitively, if the terms are't evetually very small, the the partial sums will bouce aroud too much to coverge to a limit L. This is the essece of the proof of the test. Note that this test is oly a test for divergece, ot a test for covergece. Importat Warig: The coverse of the theorem is FALSE! Eve if the terms a ted to 0, the series does't have to coverge. (We will give a example i a momet.) a

5 + coverges. l'hô We see that lim = lim =, where we used l'hôpital's rule i the middle. + Sice this is ot zero, this series diverges. (I fact, it diverges to +.) ( ) 4 + coverges. ( ) We see that lim 4 + = lim ( ) = lim + / 4 ( ), ad this last limit does ot exist. Thus, this series diverges }{{ 4 } coverges. }{{} 4 times times Notice that the terms i this series do approach zero; oetheless, we claim that this series diverges! We just group all of the equal terms together: whe we add all of the copies of that show up, we get. Thus, the partial sum icludig all terms whose deomiator is or smaller is equal to = }{{} times, sice there are such blocks of terms. Sice this goes to +, we coclude that the series diverges to +. There are two simple types of series for which we ca write dow easy summatio formulas: geometric series ad telescopig series. 7.. Geometric Series Deitio: A geometric sequece is of the form a, ar, ar,..., for some iitial value a ad some commo ratio r. We are iterested i summig the correspodig geometric series a + ar + ar + ar 3 +. Cosider the partial sum S k = a + ar + ar + ar ar k. The rs k = ar + ar + ar ar k+, ad so S k rs k = a ar k+. Therefore, S k = a rk+ is the sum of the ite geometric series ar, provided r. r Propositio: If < r <, the the iite geometric series ar a coverges to the value. If r, r the the series diverges to + if a > 0 ad to if a < 0. Fially, if r ad a 0, the the series diverges (i a oscillatory way). Proof: Write the summatio formula as S k = a r a r rk+ for r. If < r <, the the limit as of the secod term is zero, so the series coverges to a r. If r =, the the series is just a + a + a +, which diverges to + if a > 0 ad to if a < 0. If r >, the the limit of the secod term is + if a > 0 ad if a < 0. If r =, the the series is just a a + a a +, which diverges for a 0. If r <, the the secod term oscillates betwee positive ad egative, ad grows larger (i magitude) as grows, so the series diverges.

6 Example: We sum the series Example: We sum the series = = / =. 9 0 = = 9/0 /0 =. I other words, the iite repeatig decimal is actually equal to! Example: The series Example: The series = diverges to +. ( ) = diverges. 7.. Telescopig Series A telescopig series is of the form [f() f()] + [f() f(3)] + [f(3) f(4)] + = for some fuctio f(x). We ca see just by removig the paretheses ad cacellig that Thus, to compute the sum of the iite series [f() f( + )], [f() f( + )] = f() f(k + ). [f() f( + )], we simply take the limit as k of the partial sum f() f(k + ). (If the sum diverges, the so does the series.) Example: Fid the sum of the iite series +. We use partial fractio decompositio to see that + = +. [ The we ca write + = ] = + k +. As k, the sum therefore coverges to. Example: Fid the sum of the iite series ( l + ). ( We use logarithms properties to write l + ) ( ) + = l = l( + ) l(). ( The we ca write l + ) = [l( + ) l()] = l(k + ) l() = l(k + ). As k, the sum therefore diverges to. 7.3 Positive Series: Itegral Test, Compariso Tests, Ratio ad Root Tests I geeral, it is dicult (ad frequetly, impossible) to give closed-form, simple expressios for the sums of iite series.

7 We are therefore willig to settle for determiig whether or ot a give series coverges to a ite value. If we ca see that it does coverge, the usually with oly a little more eort we ca (i priciple, most of the time) compute the value to as much accuracy as we wat usig a computer. There is a fairly extesive array of series covergece tests, givig various criteria for whe a series will coverge or diverge. I this sectio, we will discuss the most commoly used tests for series whose terms are all positive. (I the ext sectio, we will treat sequeces with egative terms.) The fudametal idea behid each of these series tests is to compare the give series to somethig else that is easier to uderstad: either a itegral, or a similar (but simpler) series. Note also that if all the terms of a series are positive, either the series diverges to +, or it is bouded above. I the latter case, the Mootoe Covergece Theorem implies that the series coverges to a ite limit. Thus, we eed oly determie whether the series coverges or diverges Itegral Test Test (Itegral Test): If f(x) is a decreasig, positive fuctio, the the series if the itegral ˆ f(x) dx coverges. f() coverges if ad oly Almost always we will use this theorem to say somethig about the sum, by computig (or at the least, checkig covergece of) the itegral. The theorem is useful because sums are hard to evaluate exactly, but itegrals are ofte easier. Proof: The essece of the proof is cotaied i the followig two 'staircase' diagrams: I the picture o the left, the red regio is the area uder the curve y = f(x) o the iterval [, k+], while the blue regio is composed of k rectagles each of width, havig height f( + ) o the iterval [, + ] for each iteger k. It is easy to see that the area of the blue regio is equal k+ ˆ k+ to the sum f( + ), while the area of the red regio is equal to the itegral f(x) dx. Sice = k+ f(x) is decreasig, the blue regio is cotaied i the red regio, so f( + ) ˆ k+ f(x) dx. I the picture o the right, the red regio is the area uder the curve y = f(x) o the iterval [, k + ], while the blue regio is composed of k rectagles each of width, havig height f() o the iterval [, + ]. This time, the area of the blue regio is equal to the sum f(), while the area of the red regio is equal to the itegral cotais the red regio, so ˆ k+ f(x) dx ˆ k+ f(). k+ Combiig the two iequalities gives f( + ) = f(x) dx. Sice f(x) is decreasig, the blue regio ˆ k+ f(x) dx f(). Takig the limit as

8 [ ] [ˆ k shows that f() f() ] [ f(x) dx beig ite forces the itegral to be ite, ad vice versa. f() ]. Thus, we see that the sum We also remark that we do ot eed to start the summatio at, sice the covergece of a series is ot aected if we remove a ite umber of terms. We may replace with ay iteger ad the result will still hold. Our proof of the Itegral Test also gives us a techique for boudig the value of a series. Explicitly: Corollary: If f(x) is a decreasig positive fuctio ad S = f() is ite, the for S k = ˆ ˆ the iequality S k + f(x) dx S S k + f(x) dx. k+ k We ca rearrage this to see that S S k ˆ k f(x) dx. f() we have Hece, if we wat to estimate the value of S withi a error of ɛ, we simply eed to d the smallest k such that ˆ k f(x) dx ɛ: the S k will be withi ɛ of S. Example (p-series): For each positive real umber p, determie whether the series p coverges. We apply the Itegral Test: we thus eed to determie the covergece of the itegral is x p dx. ( x p If p < the the itegral is p sice p > 0. x p dx = ) x=, which diverges to, because x p teds to as x does, If p = the the itegral is (l(x)) x=, which diverges to, because l(x) teds to as x. ( ) x p If p > the the itegral is x= =, sice p < 0. The itegral coverges, ad therefore p p the sum coverges. Remark: The sum, the p-series with p =, is called the harmoic series. It is the simplest example of a o-coverget series whose terms oetheless shrik to zero. Example: Give a estimate for the value of the series that is accurate to (at least) two decimal places. 3 We kow ˆ this series coverges by the previous example. The corollary tells us to d the value of k such that dx 0.0. k x3 ˆ We compute x 3 dx = x x=k = k. Thus we wat to pick k such that k 0.0. k If we choose k = 8, the k = 8 < 0.0. Hece the partial sum S 8 = of the full iite sum. 8 3 = is guarateed to be withi = l() is coverget or diverget.

9 ˆ Applyig the Itegral Test idicates we should determie the covergece of the itegral Upo makig the substitutio u = l(x), with du = x dx, we obtai l(u) =, sice the atural logarithm goes to. u=l() x (l(x)) dx = l() Sice the itegral diverges to, by the Itegral Test we coclude that the sum diverges to. is coverget or diverget. e Applyig the Itegral Test idicates we should determie the covergece of the itegral ˆ x (l(x)) dx. x e x dx = u du = Now we itegrate by parts: recall that the itegratio by parts formula says f (x) g(x) dx = f(x) g(x) f(x) g (x) dx. We take f (x) = e x with f(x) = e x, ad g(x) = x with g (x) =. This yields x e x dx = x e x ( e x ) dx = x e x + e x dx = x e x e x + C. Hece ˆ via L'Hôpital's Rule. x e x dx = [ x e x e x] x= = e + e =, where we computed lim e x xe x = Sice the itegral coverges, by the Itegral Test we coclude that the sum coverges. ˆ lim x x e x = 0 x e x dx Compariso Test Test (Compariso Test): Give two sequeces {a } ad {b } of positive umbers such that a < b for all, if b coverges the so does a. Also, if a diverges the so does b. This just says that if a (positive) series coverges, the ay other series with smaller (positive) terms also coverges. Similarly, if a (positive) series diverges, the ay other series with bigger terms also diverges. The proof of the test is merely these observatios applied to the sequece of partial sums of the two series, alog with a appeal to the Mootoe Covergece Theorem. We observe that > =. Sice the sum as well. = = is coverget or diverget. diverges by the Itegral Test, the Compariso Test says that si() We observe that si() for ay, ad > 3. Hece we have si() Sice the sum coverges. is coverget or diverget. = 3 coverges by the Itegral Test, the Compariso Test says that diverges si() 3 + 5

10 7.3.3 Limit Compariso Test Test (Limit Compariso Test): Give sequeces {a } ad {b } of positive umbers, if lim costat c, the a coverges if ad oly if b coverges. a b is some positive This just says that if the terms i two series are (fairly close) to beig a positive costat times the other, the the two series either both coverge or both diverge. The proof of the test is essetially just this idea, doe carefully. I geeral, to use the Limit Compariso Test, the idea is to d a simpler series b such that lim is ite ad positive, such that b is easier to aalyze is coverget or diverget. For large, the umerator will be domiated by the term ad the deomiator will be domiated by the 3 term. Thus, we will try comparig the give series to the series with b = 3 =. For a = ad b = a ( )/( 3 + 5) 3, we have lim = lim b / = lim 3 =, by + 0 stadard limit properties (or L'Hôpital's Rule applied three times). So, because we kow that b = coverges by the Itegral Test, the Limit Compariso Test says that a also coverges. si() + l() is coverget or diverget. We will deal with the umerator ad deomiator separately by makig two comparisos. First, the umerator will be domiated by the term, because sie is betwee ad, so we begi by comparig the give series to the series with b = + l(). For a = si() + l() ad b =, we have + l() a ( si())/( ( + l()) si() lim = lim b /( = lim = lim si() ) =, + l()) where i the last step we used the squeeze theorem (sice sie is bouded as ). So we are reduced to determiig whether b = + l() coverges. We do the same procedure for this series: the deomiator will be domiated by the term, so we will compare to the series with c = =. For b = + l() ad c = b /( + l()), we have lim = lim = lim c / + l() = by a few applicatios of L'Hôpital's Rule. Fially, sice c = diverges by the Itegral Test, we coclude that a also diverges. a b

11 7.3.4 Ratio Test a + Test (Ratio Test): If the sequece {a } of positive real umbers has the property that lim exists ad a equals some costat ρ, the the sum a coverges if ρ <, ad diverges if ρ >. If ρ = the the test is icoclusive, while if ρ = the the series diverges. The idea behid the proof of the test is to compare the sequece {a } to a geometric series with commo ratio ρ. a + Proof: First suppose that the limit lim = ρ is less tha. a Let ɛ = ρ. By deitio, there exists some N such that a + ɛ for every N. (This follows because ɛ is equal to ρ + ɛ, whereas the fact that the limit lim that there caot be iitely may terms with a + a > ɛ.) a a + a is equal to ρ meas Thus we have a N+ ( ɛ)a N, a N+ ( ɛ)a N+ ( ɛ) a N, ad, by repeatig this argumet, a N+k ( ɛ) k a N. Therefore, we have the upper boud a ( ɛ) N a N = a N, sice this last sequece ɛ =N =N is a geometric series with commo ratio ɛ. We coclude that a is bouded above, so by the Mootoe Covergece Theorem it coverges. If the value of ρ is bigger tha, the we ca use essetially the same argumet to compare the series to a geometric series with commo ratio larger tha, to see that the series diverges. (We will omit the details.) e! is coverget or diverget. We see that for a = e! we have a + = e+ /( + )! a e = e+! /! e ( + )! = e a +, so lim = 0. + a Therefore the series coverges by the Ratio Test. ()! (!) is coverget or diverget. We see that for a = ()! (!) we have a + ( + )!/ [( + )!] = a ()!/[!] = ( + )( + ) ( + )( + ) a + Now we see that lim = 4, either by basic limits or two applicatios of L'Hôpital's Rule. a Therefore the series diverges by the Ratio Test. = Root Test Test (Root Test): If the sequece {a } of positive real umbers has the property that lim a exists ad equals some costat ρ, the the sum a coverges if ρ <, ad diverges if ρ >. If ρ = the test is icoclusive, while if ρ = the the series diverges. This test is like the Ratio Test but ca work better for certai types of series. The idea of the proof is the same, though: it compares the sequece {a } to a geometric series with commo ratio ρ.

12 Proof: First suppose that the limit lim a = ρ is less tha. Let ɛ = ρ. By deitio, there exists some N such that a ɛ for every N. (This follows because ɛ is equal to ρ + ɛ, whereas the fact that the limit lim a is equal to ρ meas that there caot be iitely may terms with a > ɛ.) Hece, for ay N, we have a ( ɛ). Therefore, we have the upper boud a ( ɛ) ( ɛ)n =, sice this last sequece is ɛ =N =N a geometric series with commo ratio ɛ. We coclude that a is bouded above, so by the Mootoe Covergece Theorem it coverges. If the value of ρ is bigger tha, the we ca use essetially the same argumet to compare the series to a geometric series with commo ratio larger tha, to see that the series diverges. (We will omit the details.) e is coverget or diverget. For a = e, we see that a = e, ad so lim a = 0. Therefore the series coverges by the Root Test. is coverget or diverget. For a =, we see that a = applicatio of L'Hôpital's Rule. Therefore the series diverges by the Root Test.. The lim a = lim = by basic limit properties or a 7.4 Geeral Series: Alteratig Series Test, Absolute ad Coditioal Covergece Up util ow we have dealt primarily with series whose terms are positive, but we will ow broade our aalysis to series which have both positive ad egative terms. The most commo of these are alteratig series, which are of the form with each u > 0. We will rst aalyze alteratig series, ad the broade our discussio to geeral series. ( ) u = u 0 u +u u 3 +, 7.4. Alteratig Series Test We have a special covergece test for alteratig series: Test (Alteratig Series Test): Suppose ad lim u = 0. The the series ( ) u is a alteratig series with u > u + > 0 for all, ( ) u coverges. Furthermore, if S is the value of the iite series ad S k is the kth partial sum, the S S k u k+ for every k. Note: If lim u 0, the the alteratig series diverges by our earlier results.

13 The idea behid the proof of this test is that the partial sums alterate above ad below the limit of the sum. Sice the partial sums get closer ad closer to each other, evetually they must coverge i o a sigle limitig value. Here is a illustratio of this pheomeo for the alteratig series : Proof of Test: Let S k deote the kth partial sum. First, observe that S k+ + (u k+ u k+3 ) = S k+3. Sice the term i the paretheses is positive, by the assumptio that u > u + for all, we coclude that S k+ < S k+3 for every k. Thus, S < S 3 < S 5 < S 7 <, so the odd-umbered partial sums form a icreasig sequece. I a similar way, we observe that S k (u k+ u k+ ) = S k+. Agai, sice the term i the paretheses is positive, we coclude that S k > S k+ for every k. Thus, S > S 4 > S 6 > S 7 <, so the eve-umbered partial sums form a decreasig sequece. Sice S k u k+ = S k+, we obtai the chai of iequalities S < S 3 < S 5 < < S 6 < S 4 < S. I particular, the odd-umbered partial sums form a icreasig sequece that is bouded above by S. Hece by the Mootoe Covergece Theorem, the odd-umbered partial sums coverge to a limit L. The we also have lim S k = lim S k + lim u k = L + 0 = L, so the eve-umbered k k k partial sums coverge to the same limit. For the error estimate, we simply observe that S always lies betwee S k ad S k+ for ay k, ad therefore S S k S k+ S k = u k+. ( ) + is coverget or diverget. Here u = ad we ca see that the criteria u > u + > 0 ad lim u = 0 for the Alteratig Series Test are both satised. Therefore, by the Alteratig Series Test, this series coverges. Note: This series is called the alteratig harmoic series; compare it to the (regular) harmoic series. Observe i particular that the regular harmoic series does ot coverge, but the alteratig harmoic series does coverge. Example: Show that the series ( ) + + is coverget ad estimate its value withi a error of 0.0. Here u = + ad we ca see that the criteria u > u + > 0 ad lim u = 0 for the Alteratig Series Test are both satised. Therefore, by the Alteratig Series Test, this series coverges. For the estimatio of the value, we kow that S S k u k+, so we wat to choose k so that u k+ 00 (sice this will give eough accuracy). Sice u 0 =, we ca take k = 9. 0 The desired estimate is 0.0 of this value. 9 ( ) : we are the guarateed that the iite sum is withi

14 ( ) + is coverget or diverget. Here u = + ad we ca check that u > u + > 0 by observig that u =, ad otig + / that ( + ) + + > + > +. Sice clearly lim u = 0, the Alteratig Series Test applies ad says that the series coverges. ( ) + is coverget or diverget. Here u =. We try applyig the Alteratig Series Test, but we caot, because lim + + =, by basic limit properties (or L'Hôpital's Rule). From this limit we see that the terms of the series do ot ted to zero: thus, the series actually diverges. ( ) Example: Estimate the value of the series! to four decimal places. It is easy to apply the Alteratig Series Test to see that this series coverges: we have u =!, ad the terms are clearly positive ad decrease to zero as. For the estimatio of the value, we kow that S S k u k+, so we wat to choose k so that u k+ 0 4 (sice this will give eough accuracy). Sice 8! = 4030 we see that u 8 < 0 4, so we ca take k = 7. 7 ( ) We obtai the estimate , so to four decimal places the sum is ! 7.4. Absolute ad Coditioal Covergece Sometimes alteratig series will coverge because the terms u decrease i size so rapidly that the series would have coverged eve if we summed the series without the alteratig sigs: amely, if the series ( ) u = u 0 + u + u + u 3 + is coverget. This idea is captured i the followig theorem: Test (Absolute Covergece Test): If a, a,... is a sequece of real umbers ad the series the so does the series a. The idea of the proof is to use the Compariso Test o a ad we eed to modify the rst series slightly to make its terms oegative. Proof: Suppose a coverges ad let b = a + a. a coverges, a. I order to do this, however, Notice that 0 a + a a for each, because a is either equal to a or to a. By hypothesis, the sequece a coverges hece so does the sequece a. Now because 0 b a, applyig the Compariso Test shows that b coverges.

15 Fially, sice a = b a, we see that a = b a is a dierece of two coverget series, hece is coverget. Deitio: A series which still coverges whe we take the absolute values of all the terms is said to coverge absolutely. A series which itself coverges but whose absolute-value series does ot coverge is said to coverge coditioally. The theorem above says that every absolutely coverget series coverges. I geeral, absolutely coverget series are much better behaved tha coditioally coverget series. diverget. ( ) + is absolutely coverget, coditioally coverget, or We aalyze the series of absolute values rst. Notice that this series is, ad the Itegral Test says that this series coverges. Therefore, the origial series coverges absolutely. Example: Determie whether the alteratig harmoic series is absolutely coverget, coditioally coverget, or diverget. ( ) + We aalyze the series of absolute values rst. Notice that this series is, which diverges by the Itegral Test. For the origial series, as we saw above, the Alteratig Series Test implies that the series coverges. Therefore, the series coverges coditioally Rearragemets of Iite Series Aother topic we ca ivestigate is what happes if we sum the terms of a coverget series i a dieret order. I other words: what happes if we rearrage the terms i the series? (For example, oe rearragemet of a is give by a + a 3 + a 5 + a 36 + a 5 + a 7 + a 80 +.) Based o the behavior of ite sums, it may seem that rearragig the terms caot possibly make a dierece, sice (for example) a + b + c + d is equal to d + c + a + b. However, this turs out ot to be the case: it is (quite uexpectedly) possible to chage the value of a iite order by rearragig its terms! For a explicit example, let be the alteratig harmoic series. S = Now divide each of the terms by ad pad the series by icludig zeroes: we get S =

16 Now add these two series together term-by-term. By the limit laws, the summed series is which, after removig the zero terms, yields 3 S = S = It is straightforward to verify that every term of the alteratig harmoic series occurs exactly oce. (The sequece ow has two positive terms followed by a egative term, rather tha alteratig.) But otice that we have chaged the sum by doig this rearragemet: it is ow 3 Theorem (Riema Rearragemet Theorem): Suppose coverges absolutely, the ay rearragemet of the origial value! a is a coverget series. If the series b of the series has the same sum. coverges coditioally, the there exists a rearragemet umber value, +, or. b If the series a a of the series that has ay desired real The o-ituitive behavior displayed by coditioally coverget series uderscores the fact that iite summatio ca be extremely (!) delicate. I geeral, it is of cetral importace to be scrupulously careful whe dealig with iite series, because eve somethig as seemigly iocuous as rearragig the terms ca completely chage the behavior of the series. Outlie Proof (rst part): Cosider the kth partial sum B k = b. Choose k large eough so that each of a, a,, a N appear i the sum The the dierece betwee N a. a ad b is at most Takig N larger ad larger shows that the partial sums =N+ b. a, sice the secod sum cotais b approach the value a, as claimed. Outlie Proof (secod part): First, we ote that the sum of all of the egative terms i a coditioally coverget series must be. This follows because if the sum of all the egative terms were a ite iteger N, the the series a would be equal to N + a, which coverges. (This cotradicts the assumptio that the series is ot absolutely coverget.) Similarly, the sum of all the positive terms i the series must be +. Now, i order to get a rearragemet b whose sum is r 0, we sum positive terms util the sum exceeds r, the egative terms util the sum drops below r, the more positive terms util the sum exceeds r, ad so o. The ature of the summatio will make the sum hoe i o the value r, sice each partial sum is a distace at most a t from r (where t is a parameter that icreases as we take partial sums farther out i the series), ad the terms a i shrik to zero.

17 For a egative sum, we simply sum egative terms rst. To get a sum of +, we add positive terms util the sum exceeds, the egative terms util the sum drops below, the positive terms util the sum exceeds 4, the egative terms util the sum drops below 3, ad so o ad so forth. Ad for, we simply iterchage positive ad egative. 7.5 Further Examples of Series Covergece Tests I this sectio, we give a umber of examples of series ad apply the various series tests to determie absolute / coditioal covergece of the correspodig series. We have icluded these examples i a separate sectio to give additioal practice for determiig which tests to use o dieret types of iite series. Here is a geeral list of steps to follow whe tryig to determie the covergece of a give series a : First, determie lim a : if this limit fails to exist, or exists but is ozero, the series diverges. Next, examie the terms i the series. There are dieret strategies depedig o the form of the terms of the series: Suppose rst that the terms of the series are positive: If the series is a geometric series of the form ar, it ca be summed directly. Also cosider the possibility that the series may be a telescopig series of the form [f() f( + )] for some ice fuctio f(x), or a sum of several telescopig or geometric series. If a = q() where q() is a ratioal fuctio (or more geerally, a algebraic fuctio, possibly ivolvig radicals), use the Limit Compariso Test to compare the series to a appropriate p-series. p More geerally, if the terms of a ca be easily compared to a simpler fuctio, use the Limit Compariso Test to covert the problem to oe about aalyzig a simpler series. If a = f() where f(x) is a ice fuctio (whose itegral ˆ f(x) dx is easily evaluated), use the Itegral Test. If a = (b ) where b is simple, the try usig the Root Test. Particularly worth otig is the fact that lim = (which may be derived from L'Hôpital's Rule). If a ivolves factorials, expoetials, or other kids of products, try usig the Ratio Test. (Note that usig the Ratio Test o quotiets of polyomials will ever work, because the quotiet q( + )/q() of successive terms always teds to for ay ratioal fuctio q(x).) Now suppose that the series has egative terms: First aalyze the absolute value series a usig the tests above: if this series coverges, the so does the origial series. If the series is a alteratig series of the form ( ) u, try usig the Alteratig Series Test. I some cases, it may be ecessary to make algebraic maipulatios to simplify the terms of the series before applyig ay of the tests. diverget. ( ) is absolutely coverget, coditioally coverget, or We compute lim a 3 = lim =. Sice this is ozero, we coclude that the series is diverget. 3 +

18 ( 3) 3! diverget. is absolutely coverget, coditioally coverget, or We examie lim a 3 3 = lim. This limit will be zero because factorials grow faster tha expoetials ad polyomial fuctios.! 3 3 Now we aalyze the absolute value series, with b = 3 3.!! Sice the terms b = 3 3! ivolve factorials ad expoetials, we try usig the Ratio Test. We compute b + = 3+ ( + ) 3 /( + )! b 3 3 = 3+ ( + ) 3! /! 3 3 = ( + )! 3( + ) 3. As, we see that b + 0, sice the deomiator has a higher degree tha the umerator. b Hece by the Ratio Test, the absolute value series coverges. We coclude that the origial series is absolutely coverget.! is absolutely coverget, coditioally coverget, or diverget. Sice the terms of this series are positive, we oly eed to determie covergece. It is ot immediate whether lim a! = lim coverges to zero or ot. Sice the terms a =! ivolve factorials ad expoetials, we try usig the Ratio Test. We compute a + a = As, we see that a + a ( + )!/(+)!/ = ( + )!! = , sice the expoetial i the deomiator will domiate the umerator. Hece by the Ratio Test, the series is (absolutely) coverget. is coverget or diverget. +/ =3 It is easy to see that lim a = lim coverges to zero, because the deomiator is bigger tha +/ =. The fuctio f(x) = is ot easy to itegrate. It is also ot easy to take the th root of the terms, x+/x or is the ratio a + particularly ice. a We do otice that the terms are rather similar to the terms of the harmoic series. We will try =3 usig the Limit Compariso Test with a = ad b =. +/ We see that a b = / +/ = /+/ =. We kow that lim =, so the Limit Compariso Test says that our series is diverget, because we kow that the harmoic series is diverget. More specically, it diverges to +.

19 or diverget. 3 ( ) is absolutely coverget, coditioally coverget, We examie lim a = lim while the umerator grows like / Sice the terms a = By basic limit properties, lim the same rate as But we kow that 3 3 = / This limit will be zero because the deomiator grows like 3/ ivolve algebraic fuctios, we try usig the Limit Compariso Test =, ad lim =, so the terms a grow at essetially 3 a =. 5/6 = 3/ 5/6. Explicitly, by the limit laws, we deduce that lim 5/6 diverges to sice it is a p-series. Hece the origial series is ot absolutely coverget by the Limit Compariso Test. Now we examie the origial series, which is a alteratig series. 3 + We apply the Alteratig Series Test with u =. It is a straightforward check that f(x) = is a decreasig fuctio for large eough, ad as we saw above, lim u = 0. Hece by the Alteratig Series Test, the origial series coverges. We coclude that the origial series is coditioally coverget. Clearly, lim a = 0. =3 l() is coverget or diverget. The th root of a is ot especially ice, or is a +, so the Root ad Ratio Tests are ulikely to be useful. The oly atural series for compariso is the harmoic series, but the limit of the ratio a / = is zero as. l() ˆ We try the Itegral Test: we compute 3 x l(x) dx. We substitute u = l(x) with du = x dx to obtai du = u / l(3) u u=l(3) =. Hece, by the Itegral Test, we coclude that the origial series is diverget. More specically, it diverges to +. ( ) [ + ] is absolutely coverget, coditioally coverget, or diverget. a For this series, we rst rewrite the terms usig a b = Agai, we start by aalyzig the absolute value series with a = (or Limit Compariso) Test, the series a b ( ) : this gives a =. a + b By the Compariso a diverges, because we ca compare it to the series with b =, which is a diverget p-series. b

20 For the origial series, we apply the Alteratig Series Test. Each of the hypotheses is easy to verify (we skip the details), ad so the test implies that the series coverges. We coclude that the origial series is coditioally coverget. ( ) [ l( + 3) l() ] is absolutely coverget, coditioally coverget, or diverget. For this series, we rst combie ( the logarithms ito a sigle term by writig l( + 3) l() = l( + 3) l( ) + 3 ) = l = l ( + 3 ). Now we compute lim a = lim ad f(x) = l(x) is cotiuous. l ( + 3 ) = l(), sice the term iside the logarithm teds to Sice the limit is ot zero, we coclude that the origial series is diverget. cos(3) + is absolutely coverget, coditioally coverget, or diverget. We compare the absolute value series to a simpler oe. First, we observe that cos(3) +, because cosie is always betwee ad. + Thus, by the Compariso Test, if the secod series coverges, the so does the rst oe. Now we use the Limit Compariso Test to compare /( ) ratio is lim /( + ) = lim + that + coverges. + with = cos(3) + : sice the limit of the =, ad the secod series is a coverget p-series, we see Thus, by the Compariso Test, the origial series is absolutely coverget Here, we observe that 3 4, so a is coverget or diverget. The we apply the Root Test to see that lim a lim Sice this limit is less tha, the series coverges. I fact, if we rewrite the series as ad that the exact value is ( 3 5 ) + 3/5 + 4/5 = 5. 4 = ( ) 4, we see that it is the sum of two geometric series, 5 Well, you're at the ed of my hadout. Hope it was helpful. Copyright otice: This material is copyright Eva Dummit, You may ot reproduce or distribute this material without my express permissio.