INFINITE SERIES KEITH CONRAD


 Dennis Neil Reynolds
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1 INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal limit process: ifiite series. The label series is just aother ame for a sum. A ifiite series is a sum with ifiitely may terms, such as (.) The idea of a ifiite series is familiar from decimal expasios, for istace the expasio ca be writte as π = π = , so π is a ifiite sum of fractios. Decimal expasios like this show that a ifiite series is ot a paradoxical idea, although it may ot be clear how to deal with odecimal ifiite series like (.) at the momet. Ifiite series provide two coceptual isights ito the ature of the basic fuctios met i high school (ratioal fuctios, trigoometric ad iverse trigoometric fuctios, expoetial ad logarithmic fuctios). First of all, these fuctios ca be expressed i terms of ifiite series, ad i this way all these fuctios ca be approximated by polyomials, which are the simplest kids of fuctios. That simpler fuctios ca be used as approximatios to more complicated fuctios lies behid the method which calculators ad computers use to calculate approximate values of fuctios. The secod isight we will have usig ifiite series is the close relatioship betwee fuctios which seem at first to be quite differet, such as expoetial ad trigoometric fuctios. Two other applicatios we will meet are a proof by calculus that there are ifiitely may primes ad a proof that e is irratioal. 2. Defiitios ad basic examples Before discussig ifiite series we discuss fiite oes. A fiite series is a sum a + a 2 + a a N, where the a i s are real umbers. I terms of Σotatio, we write N a + a 2 + a a N = a. =
2 2 KEITH CONRAD For example, (2.) N = = N. The sum i (2.) is called a harmoic sum, for istace = A very importat class of fiite series, more importat tha the harmoic oes, are the geometric series N (2.2) + r + r r N = r. A example is = ( ) = 2 = = 2 = = The geometric series (2.2) ca be summed up exactly, as follows. Theorem 2.. Whe r, the series (2.2) is Proof. Let The + r + r r N = rn+ r S N = + r + + r N. rs N = r + r r N+. = rn+ r. These sums overlap i r + + r N, so subtractig rs N from S N gives ( r)s N = r N+. Whe r we ca divide ad get the idicated formula for S N. Example 2.2. For ay N, Example 2.3. For ay N, N = 2N N = (/2)N+ /2 = 2 N+. = 2 2 N. It is sometimes useful to adjust the idexig o a sum, for istace (2.3) = 2 = = 99 = ( + )2.
3 INFINITE SERIES 3 Comparig the two Σ s i (2.3), otice how the reumberig of the idices affects the expressio beig summed: if we subtract from the bouds of summatio o the Σ the we add to the idex o the iside to maitai the same overall sum. As aother example, = 2 = ( + 3) 2. =3 Wheever the idex is shifted dow (or up) by a certai amout o the Σ it has to be shifted up (or dow) by a compesatig amout iside the expressio beig summed so that we are still addig the same umbers. This is aalogous to the effect of a additive chage of variables i a itegral, e.g., (x + 5) 2 dx = 6 5 = u 2 du = 6 5 x 2 dx, where u = x + 5 (ad du = dx). Whe the variable iside the itegral is chaged by a additive costat, the bouds of itegratio are chaged additively i the opposite directio. Now we defie the meaig of ifiite series, such as (.). The basic idea is that we look at the sum of the first N terms, called a partial sum, ad see what happes i the limit as N. Defiitio 2.4. Let a, a 2, a 3,... be a ifiite sequece of real umbers. a is defied to be N a = lim a. N = The ifiite series If the limit exists i R the we say a is coverget. If the limit does ot exist or is ± the a is called diverget. Notice that we are ot really addig up all the terms i a ifiite series at oce. We oly add up a fiite umber of the terms ad the see how thigs behave i the limit as the (fiite) umber of terms teds to ifiity: a ifiite series is defied to be the limit of its sequece of partial sums. Example 2.5. Usig Example 2.3, So 2 = lim N N = is a coverget series ad its value is = lim N 2 2 N = 2. Buildig o this example we ca compute exactly the value of ay ifiite geometric series. Theorem 2.6. For x R, the (ifiite) geometric series x coverges if x < ad diverges if x. If x < the x = x. Proof. For N, Theorem 2. tells us { N x ( x N+ )/( x), if x, = N +, if x =. =
4 4 KEITH CONRAD Whe x <, x N+ as N, so N x = lim N = x x N+ = lim N x = x. Whe x > the umerator x N does ot coverge as N, so x diverges. (Specifically, the limit is if x > ad the partials sums oscillate betwee values tedig to ad if x <.) What if x =? If x = the the Nth partial sum is N + so the series diverges (to ). If x = the N = ( ) oscillates betwee ad, so agai the series ( ) is diverget. We will see that Theorem 2.6 is the fudametal example of a ifiite series. May importat ifiite series will be aalyzed by comparig them to a geometric series (for a suitable choice of x). If we start summig a geometric series ot at, but at a higher power of x, the we ca still get a simple closed formula for the series, as follows. Corollary 2.7. If x < ad m the x = x m + x m+ + x m+2 + = m Proof. The Nth partial sum (for N m) is xm x. x m + x m+ + + x N = x m ( + x + + x N m ). As N, the sum iside the paretheses becomes the stadard geometric series, whose value is /( x). Multiplyig by x m gives the asserted value. Example 2.8. If x < the x + x 2 + x 3 + = x/( x). A diverget geometric series ca diverge i differet ways: the partial sums may ted to or ted to both ad or oscillate betwee ad. The label diverget series does ot always mea the partial sums ted to. All diverget meas is ot coverget. Of course i a particular case we may kow the partial sums do ted to, ad the we would say the series diverges to. The covergece of a series is determied by the behavior of the terms a for large. If we chage (or omit) ay iitial set of terms i a series we do ot chage the covergece of divergece of the series. Here is the most basic geeral feature of coverget ifiite series. Theorem 2.9. If the series a coverges the a as. Proof. Let S N = a + a a N ad let S = a be the limit of the partial sums S N as N. The as N, a N = S N S N S S =. While Theorem 2.9 is formulated for coverget series, its mai importace is as a divergece test : if the geeral term i a ifiite series does ot ted to the the series diverges. For example, Theorem 2.9 gives aother reaso that a geometric series x diverges if x, because i this case x does ot ted to : x = x for all.
5 INFINITE SERIES 5 It is a ufortuate fact of life that the coverse of Theorem 2.9 is geerally false: if a we have o guaratee that a coverges. Here is the stadard illustratio. Example 2.. Cosider the harmoic series turs out that =.. Although the geeral term teds to it To show this we will compare N = / with N dt/t = log N. Whe t, /t /. Itegratig this iequality from to + gives + dt/t /, so N = N = + dt t = N+ dt t = log(n + ). Therefore the Nth partial sum of the harmoic series is bouded below by log(n + ). Sice log(n + ) as N, so does the Nth harmoic sum, so the harmoic series diverges. We will see later that the lower boud log(n + ) for the Nth harmoic sum is rather sharp, so the harmoic series diverges slowly sice the logarithm fuctio diverges slowly. The divergece of the harmoic series is ot just a couterexample to the coverse of Theorem 2.9, but ca be exploited i other cotexts. Let s use it to prove somethig about prime umbers! Theorem 2.. There are ifiitely may prime umbers. Proof. (Euler) We will argue by cotradictio: assumig there are oly fiitely may prime umbers we will cotradict the divergece of the harmoic series. Cosider the product of /( /p) as p rus through all prime umbers. Sums are deoted with a Σ ad products are deoted with a Π (capital pi), so our product is writte as p /p = /2 /3 /5. Sice we assume there are fiitely may primes, this is a fiite product. Now expad each factor ito a geometric series: /p = ( + p + p 2 + p ) 3 +. p p Each geometric series is greater tha ay of its partial sums. Pick ay iteger N 2 ad trucate each geometric series at the Nth term, givig the iequality (2.4) /p > ( + p + p 2 + p ) p N. p p O the right side of (2.4) we have a fiite product of fiite sums, so we ca compute this usig the distributive law ( super FOIL i the termiology of high school algebra). That is, take oe term from each factor, multiply them together, ad add up all these products. What umbers do we get i this way o the right side of (2.4)? A product of reciprocal itegers is a reciprocal iteger, so we obtai a sum of / as rus over certai positive itegers. Specifically, the s we meet are those whose prime factorizatio does t
6 6 KEITH CONRAD ivolve ay prime with expoet greater tha N. Ideed, for ay positive iteger >, if we factor ito primes as = p e pe 2 2 per r the e i < for all i. (Sice p i 2, 2 e i, ad 2 m > m for ay m so > e i.) Thus if N ay of the prime factors of appear i with expoet less tha N, so / is a umber which shows up whe multiplyig out (2.4). Here we eed to kow we are workig with a product over all primes. Sice / occurs i the expasio of the right side of (2.4) for N ad all terms i the expasio are positive, we have the lower boud estimate (2.5) p N /p >. Sice the left side is idepedet of N while the right side teds to as N, this iequality is a cotradictio so there must be ifiitely may primes. We ed this sectio with some algebraic properties of coverget series: termwise additio ad scalig. Theorem 2.2. If a ad b coverge the (a + b ) coverges ad + b ) = (a a + b. If a coverges the for ay umber c the series ca coverges ad a. = ca = c Proof. Let S N = N = a ad T N = N = b. Set S = a ad T = b, so S N S ad T N T as N. The by kow properties of limits of sequeces, S N + T N S + T ad cs N cs as N. These are the coclusios of the theorem. Example 2.3. We have ad ( 2 + ) 3 = /2 + /3 = = 5 4 = 5 /4 = Positive series I this sectio we describe several ways to show that a ifiite series a coverges whe a > for all. There is somethig special about series where all the terms are positive. What is it? The sequece of partial sums is icreasig: S N+ = S N + a N, so S N+ > S N for every N. A icreasig sequece has two possible limitig behaviors: it coverges or it teds to. (That is, divergece ca oly happe i oe way: the partial sums ted to.) We ca tell if a icreasig sequece coverges by showig it is bouded above. The it coverges ad its limit is a upper boud o the whole sequece. If the terms i a icreasig sequece are ot bouded above the the sequece
7 INFINITE SERIES 7 is ubouded ad teds to. Compare this to the sequece of partial sums N = ( ), which are bouded but do ot coverge (they oscillate betwee ad ). Let s summarize this property of positive series. If a > for every ad there is a costat C such that all partial sums satisfy N = a < C the a coverges ad a C. O the other had, if there is o upper boud o the partial sums the a =. We will use this over ad over agai to explai various covergece tests for positive series. Theorem 3. (Itegral Test). Suppose a = f() where f is a cotiuous decreasig positive fuctio. The a coverges if ad oly if f(x) dx coverges. Proof. We will compare the partial sum N = a ad the partial itegral N f(x) dx. Sice f(x) is a decreasig fuctio, x + = a + = f( + ) f(x) f() = a. Itegratig these iequalities betwee ad + gives (3.) a + + sice the iterval [, + ] has legth. Therefore (3.2) ad (3.3) N a = N a = a + = Combiiig (3.2) ad (3.3), (3.4) N = + N a a + =2 N+ f(x) dx N =2 N = f(x) dx a. f(x) dx = N+ f(x) dx N f(x) dx = f() + f(x) dx. N a f() + f(x) dx. Assume that f(x) dx coverges. Sice f(x) is a positive fuctio, N f(x) dx < f(x) dx. Thus by the secod iequality i (3.4) N a < f() + = f(x) dx, so the partial sums are bouded above. Therefore the series a coverges. Coversely, assume a coverges. The the partial sums N = a are bouded above (by the whole series a ). Every partial itegral of f(x) dx is bouded above by oe of these partial sums accordig to the first iequality of (3.4), so the partial itegrals are bouded above ad thus the improper itegral f(x) dx coverges. Example 3.2. We were implicitly usig the itegral test with f(x) = /x whe we proved the harmoic series diverged i Example 2.. For the harmoic series, (3.4) becomes (3.5) log(n + ) N = + log N.
8 8 KEITH CONRAD Thus the harmoic series diverges logarithmically. The followig table illustrates these bouds whe N is a power of. N log(n + ) N = / + log N Table Example 3.3. Geeralizig the harmoic series, cosider p = + 2 p + 3 p + 4 p + where p >. This is called a pseries. Its covergece is related to the itegral dx/x p. Sice the itegrad /x p has a atiderivative that ca be computed explicitly, we ca compute the improper itegral: { dx x p = p, if p >,, if < p. Therefore /p coverges whe p > ad diverges whe < p. For example, /2 coverges ad / diverges. Example 3.4. The series 2 log 2 diverges by the itegral test sice dx x log x = log log x 2 =. (We stated the itegral test with a lower boud of but it obviously adapts to series where the first idex of summatio is somethig other tha, as i this example.) Example 3.5. The series, with terms just slightly smaller tha i the previous (log ) 2 2 example, coverges. Usig the itegral test, dx x(log x) = log x = log 2, which is fiite. 2 Because the covergece of a positive series is itimately tied up with the boudedess of its partial sums, we ca determie the covergece or divergece of oe positive series from aother if we have a iequality i oe directio betwee the terms of the two series. This leads to the followig covergece test. Theorem 3.6 (Compariso test). Let < a b for all. If b coverges the a coverges. If a diverges the b diverges. 2
9 INFINITE SERIES 9 Proof. Sice a b, (3.6) N a = N b. If b coverges the the partial sums o the right side of (3.6) are bouded, so the partial sums o the left side are bouded, ad therefore a coverges. If a diverges the N = a, so N = b, so b diverges. Example 3.7. If x < the x / coverges sice x / x so the series is bouded above by the geometric series x, which coverges. Notice that, ulike the geometric series, we are ot givig ay kid of closed form expressio for x /. All we are sayig is that the series coverges. The series x / does ot coverge for x : if x = it is the harmoic series ad if x > the the geeral term x / teds to (ot ) so the series diverges to. Example 3.8. The itegral test is a special case of the compariso test, where we compare a with + f(x) dx ad with f(x) dx. Just as covergece of a series does ot deped o the behavior of the iitial terms of the series, what is importat i the compariso test is a iequality a b for large. If this iequality happes ot to hold for small but does hold for all large the the compariso test still works. The followig two examples use this exteded compariso test whe the form of the compariso test i Theorem 3.6 does ot strictly apply (because the iequality a b does t hold for small ). Example 3.9. If x < the x coverges. This is obvious for x =. We ca t get covergece for < x < by a compariso with x because the iequality goes the wrog way: x > x, so covergece of x does t help us show covergece of x. However, let s write = x = x /2 x /2. Sice < x <, x /2 as, so x /2 for large (depedig o the specific value of x, e.g, for x = /2 the iequality is false at = 3 but is correct for 4). Thus for all large we have x x /2 so x coverges by compariso with the geometric series x/2 = ( x). We have ot obtaied ay kid of closed formula for x, however. If x the x diverges sice x for all, so the geeral term does ot ted to. Example 3.. If x the the series x /! coverges. This is obvious for x =. For x > we will use the compariso test ad the lower boud estimate! > /e. Let s recall that this lower boud o! comes from Euler s itegral formula The for positive x (3.7)! = t e t dt > t e t dt > x! < x ( ex ) /e =. e t dt = e.
10 KEITH CONRAD Here x is fixed ad we should thik about what happes as grows. The ratio ex/ teds to as, so for large we have ex (3.8) < 2 (ay positive umber less tha will do here; we just use /2 for cocreteess). Raisig both sides of (3.8) to the th power, ( ex ) < 2. Comparig this to (3.7) gives x /! < /2 for large, so the terms of the series x /! drop off faster tha the terms of the coverget geometric series /2 whe we go out far eough. The exteded compariso test ow implies covergece of x /! if x >. Example 3.. The series Example coverges sice < 5 2 ad coverges by 52 Let s try to apply the compariso test to /(52 ). The rate of growth of the th term is like /(5 2 ), but the iequality ow goes the wrog way: 5 2 < 5 2 for all. So we ca t quite use the compariso test to show /(52 ) coverges (which it does). The followig limitig form of the compariso test will help here. Theorem 3.2 (Limit compariso test). Let a, b > ad suppose the ratio a /b has a positive limit. The a coverges if ad oly if b coverges. Proof. Let L = lim a /b. For all large, so L 2 < a b < 2L, L (3.9) 2 b < a < 2Lb for large. Sice L/2 ad 2L are positive, the covergece of b is the same as covergece of (L/2)b ad (2L)b. Therefore the first iequality i (3.9) ad the compariso test show covergece of a implies covergece of b. The secod iequality i (3.9) ad the compariso test show covergece of b implies covergece of a. (Although (3.9) may ot be true for all, it is true for large, ad that suffices to apply the exteded compariso test.) The way to use the limit compariso test is to replace terms i a series by simpler terms which grow at the same rate (at least up to a scalig factor). The aalyze the series with those simpler terms. Example 3.3. We retur to 5 2. As, the th term grows like /52 : /(5 2 ) /5 2.
11 INFINITE SERIES Sice /(52 ) coverges, so does /(52 ). I fact, if a is ay sequece whose growth is like / 2 up to a scalig factor the a coverges. I particular, this is a simpler way to settle the covergece i Example 3. tha by usig the iequalities i Example 3.. Example 3.4. To determie if coverges we replace the th term with the expressio = 2 3/2, which grows at the same rate. The series /(23/2 ) coverges, so the origial series coverges. The limit compariso test uderlies a importat poit: the covergece of a series a is ot assured just by kowig if a, but it is assured if a rapidly eough. That is, the rate at which a teds to is ofte (but ot always) a meas of explaiig the covergece of a. The differece betwee someoe who has a ituitive feelig for covergece of series ad someoe who ca oly determie covergece o a casebycase basis usig memorized rules i a mechaical way is probably idicated by how well the perso uderstads the coectio betwee covergece of a series ad rates of growth (really, decay) of the terms i the series. Armed with the covergece of a few basic series (like geometric series ad pseries), the covergece of most other series ecoutered i a calculus course ca be determied by the limit compariso test if you uderstad well the rates of growth of the stadard fuctios. Oe exceptio is if the series has a log term, i which case it might be useful to apply the itegral test. 4. Series with mixed sigs All our previous covergece tests apply to series with positive terms. They also apply to series whose terms are evetually positive (just omit ay iitial egative terms to get a positive series) or evetually egative (egate the series, which does t chage the covergece status, ad ow we re reduced to a evetually positive series). What do ca we do for series whose terms are ot evetually all positive or evetually all egative? These are the series with ifiitely may positive terms ad ifiitely may egative terms. Example 4.. Cosider the alteratig harmoic series ( ) = The usual harmoic series diverges, but the alteratig sigs might have differet behavior. The followig table collects some of the partial sums. N N = ( ) / Table 2
12 2 KEITH CONRAD These partial sums are ot clear evidece i favor of covergece: after, terms we oly get apparet stability i the first 2 decimal digits. The harmoic series diverges slowly, so perhaps the alteratig harmoic series diverges too (ad slower). Defiitio 4.2. A ifiite series where the terms alterate i sig is called a alteratig series. A alteratig series startig with a positive term ca be writte as ( ) a where a > for all. If the first term is egative the a reasoable geeral otatio is ( ) a where a > for all. Obviously egatig a alteratig series of oe type turs it ito the other type. Theorem 4.3 (Alteratig series test). A alteratig series whose th term i absolute value teds mootoically to as is a coverget series. Proof. We will work with alteratig series which have a iitial positive term, so of the form ( ) a. Sice successive partial sums alterate addig ad subtractig a amout which, i absolute value, steadily teds to, the relatio betwee the partial sums is idicated by the followig iequalities: s 2 < s 4 < s 6 < < s 5 < s 3 < s. I particular, the eveidexed partial sums are a icreasig sequece which is bouded above (by s, say), so the partial sums s 2m coverge. Sice s 2m+ s 2m = a 2m+, the oddidexed partial sums coverge to the same limit as the eveidexed partial sums, so the sequece of all partial sums has a sigle limit, which meas ( ) a coverges. Example 4.4. The alteratig harmoic series ( ) coverges. Example 4.5. While the pseries /p coverges oly for p >, the alteratig pseries ( ) / p coverges for the wider rage of expoets p > sice it is a alteratig series satisfyig the hypotheses of the alteratig series test. Example 4.6. We saw x / coverges for x < (compariso to the geometric series) i Example 3.7. If x < the x / coverges by the alteratig series test: the egativity of x makes the terms x / alterate i sig. To apply the alteratig series test we eed x + + < x for all, which is equivalet after some algebra to x < + for all, ad this is true sice x < ( + )/. Example 4.7. I Example 3. the series x /! was see to coverge for x. If x < the series is a alteratig series. Does it fit the hypotheses of the alteratig series test? We eed x + ( + )! < x! for all, which is equivalet to x < +
13 INFINITE SERIES 3 for all. For each particular x < this may ot be true for all (if x = 3 it fails for = ad 2), but it is defiitely true for all sufficietly large. Therefore the terms of x /! are a evetually alteratig series. Sice what matters for covergece is the logterm behavior ad ot the iitial terms, we ca apply the alteratig series test to see x /! coverges by just igorig a iitial piece of the series. Whe a series has ifiitely may positive ad egative terms which are ot strictly alteratig, covergece may ot be easy to check. For example, the trigoometric series si(x) = si x + si(2x) 2 + si(3x) 3 + si(4x) 4 turs out to be coverget for every umber x, but the arragemet of positive ad egative terms i this series ca be quite subtle i terms of x. The proof that this series coverges uses the techique of summatio by parts, which wo t be discussed here. While it is geerally a delicate matter to show a oalteratig series with mixed sigs coverges, there is oe useful property such series might have which does imply covergece. It is this: if the series with all terms made positive coverges the so does the origial series. Let s give this idea a ame ad the look at some examples. + Defiitio 4.8. A series a is absolutely coverget if a coverges. Do t cofuse a with a ; absolute covergece refers to covergece if we drop the sigs from the terms i the series, ot from the series overall. The differece betwee a ad a is at most a sigle sig, while there is a substatial differece betwee a ad a if the a s have may mixed sigs; that is the differece which absolute covergece ivolves. Example 4.9. We saw x / coverges if x < i Example 3.7. Sice x / = x /, the series x / is absolutely coverget if x <. Note the alteratig harmoic series ( ) / is ot absolutely coverget, however. Example 4.. I Example 3. we saw x /! coverges for all x. Sice x /! = x /!, the series x /! is absolutely coverget for all x. Example 4.. By Example 3.9, x coverges absolutely if x <. Example 4.2. The two series si(x) 2, cos(x) 2, where the th term has deomiator 2, are absolutely coverget for ay x sice the absolute value of the th term is at most / 2 ad /2 coverges. The relevace of absolute covergece is twofold: ) we ca use the may covergece tests for positive series to determie if a series is absolutely coverget, ad 2) absolute covergece implies ordiary covergece. We just illustrated the first poit several times. Let s show the secod poit. Theorem 4.3 (Absolute covergece test). Every absolutely coverget series is a coverget series.
14 4 KEITH CONRAD Proof. For all we have so a a a, a + a 2 a. Sice a coverges, so does 2 a (to twice the value). Therefore by the compariso test we obtai covergece of (a + a ). Subtractig a from this shows a coverges. Thus the series x / ad x coverge for x < ad x /! coverges for all x by our treatmet of these series for positive x aloe i Sectio 3. The argumet we gave i Examples 4.6 ad 4.7 for the covergece of x / ad x /! whe x <, usig the alteratig series test, ca be avoided. (But we do eed the alteratig series test to show x / coverges at x =.) A series which coverges but is ot absolutely coverget is called coditioally coverget. A example of a coditioally coverget series is the alteratig harmoic series ( ) /. The distictio betwee absolutely coverget series ad coditioally coverget series might seem kid of mior, sice it ivolves whether or ot a particular covergece test (the absolute covergece test) works o that series. But the distictio is actually profoud, ad is icely illustrated by the followig example of Dirichlet (837). Example 4.4. Let L = /2 + /3 /4 + /5 /6 + be the alteratig harmoic series. Cosider the rearragemet of the terms where we follow a positive term by two egative terms: If we add each positive term to the egative term followig it, we obtai , which is precisely half the alteratig harmoic series (multiply L by /2 ad see what you get termwise). Therefore, by rearragig the terms of the alteratig harmoic series we obtaied a ew series whose sum is (/2)L istead of L. If (/2)L = L, does t that mea = 2? What happeed? This example shows additio of terms i a ifiite series is ot geerally commutative, ulike with fiite series. I fact, this feature is characteristic of the coditioally coverget series, as see i the followig amazig theorem. Theorem 4.5 (Riema, 854). If a ifiite series is coditioally coverget the it ca be rearraged to sum up to ay desired value. If a ifiite series is absolutely coverget the all of its rearragemets coverge to the same sum. For istace, the alteratig harmoic series ca be rearraged to sum up to, to 5.673, to π, or to ay other umber you wish. That we rearraged it i Example 4.4 to sum up to half its usual value was special oly i the sese that we could make the rearragemet achievig that effect quite explicit. Theorem 4.5, whose proof we omit, is called Riema s rearragemet theorem. It idicates that absolutely coverget series are better behaved tha coditioally coverget oes. The ordiary rules of algebra for fiite series geerally exted to absolutely coverget series but ot to coditioally coverget series.
15 INFINITE SERIES 5 5. Power series The simplest fuctios are polyomials: c + c x + + c d x d. I differetial calculus we lear how to locally approximate may fuctios by liear fuctios (the taget lie approximatio). We ow itroduce a idea which will let us give a exact represetatio of fuctios i terms of polyomials of ifiite degree. The precise ame for this idea is a power series. Defiitio 5.. A power series is a fuctio of the form f(x) = c x. For istace, a polyomial is a power series where c = for large. The geometric series x is a power series. It oly coverges for x <. We have also met other power series, like x, x x,!. These three series coverge o the respective itervals [, ), (, ) ad (, ). For ay power series c x it is a basic task to fid those x where the series coverges. It turs out i geeral, as i all of our examples, that a power series coverges o a iterval. Let s see why. Theorem 5.2. If a power series c x coverges at x = b the it coverges absolutely for x < b. Proof. Sice c b coverges the geeral term teds to, so c b for large. If x < b the c x = c b x, b which is at most x/b for large. The series x/b coverges sice it s a geometric series ad x/b <. Therefore by the (exteded) compariso test c x coverges, so c x is absolutely coverget. Example 5.3. If a power series coverges at 7 the it coverges o the iterval ( 7, 7). We ca t say for sure how it coverges at 7. Example 5.4. If a power series diverges at 7 the it diverges for x > 7: if it were to coverge at a umber x where x > 7 the it would coverge i the iterval ( x, x ) so i particular at 7, a cotradictio. What these two examples illustrate, as cosequeces of Theorem 5.2, is that the values of x where a power series c x coverges is a iterval cetered at, so of the form ( r, r), [ r, r), ( r, r], [ r, r] for some r. We call r the radius of covergece of the power series. The oly differece betwee these differet itervals is the presece or absece of the edpoits. All possible types of iterval of covergece ca occur: x has iterval of covergece (, ), x / has iterval of covergece [, ), ( ) x / has iterval of covergece (, ] ad x / 2 (a ew example we have ot met before) has iterval of covergece [, ]. The radius of covergece ad the iterval of covergece are closely related but should ot be cofused. The iterval is the actual set where the power series coverges. The radius is simply the halflegth of this set (ad does t tell us whether or ot the edpoits are icluded). If we do t care about covergece behavior o the boudary of the iterval of covergece tha we ca get by just kowig the radius of covergece: the series always coverges (absolutely) o the iside of the iterval of covergece.
16 6 KEITH CONRAD For the practical computatio of the radius of covergece i basic examples it is coveiet to use a ew covergece test for positive series. Theorem 5.5 (Ratio test). If a > for all, assume a + q = lim a exists. If q < the a coverges. If q > the a diverges. If q = the o coclusio ca be draw. Proof. Assume q <. Pick s betwee q ad : q < s <. Sice a + /a q, we have a + /a < s for all large, say for. The a + < sa whe, so a + < sa, a +2 < sa + < s 2 a, a +3 < sa +2 < s 3 a, ad more geerally a +k < s k a for ay k. Writig + k as we have = a < s a = a s s. Therefore whe the umber a is bouded above by a costat multiple of s. Hece a coverges by compariso to a costat multiple of the coverget geometric series s. I the other directio, if q > the pick s with < s < q. A argumet similar to the previous oe shows a grows at least as quickly as a costat multiple of s, but this time s diverges sice s >. So a diverges too. Whe q = we ca t make a defiite coclusio sice both / ad /2 have q =. Example 5.6. We give a proof that x has radius of covergece which is shorter tha Examples 3.9 ad 4.. Take a = x = x. The a + = + x x a as. Thus if x < the series x is absolutely coverget by the ratio test. If x > this series is diverget. Notice the ratio test does ot tell us what happes to x whe x = ; we have to check x = ad x = idividually. Example 5.7. A similar argumet shows x / has radius of covergece sice as. x + /( + ) x / = x x + Example 5.8. We show the series x /! coverges absolutely for all x. Usig the ratio test we look at x + /( + )! x /! = x + as. Sice this limit is less tha for all x, we are doe by the ratio test. The radius of covergece is ifiite. Power series are importat for two reasos: they give us much greater flexibility to defie ew kids of fuctios ad may stadard fuctios ca be expressed i terms of a power series.
17 INFINITE SERIES 7 Ituitively, if a kow fuctio f(x) has a power series represetatio o some iterval aroud, say f(x) = c x = c + c x + c 2 x 2 + c 3 x 3 + c 4 x 4 + for x < r, the we ca guess a formula for c i terms of the behavior of f(x). To begi we have f() = c. Now we thik about power series as somethig like ifiite degree polyomials. We kow how to differetiate a polyomial by differetiatig each power of x separately. Let s assume this works for power series too i the iterval of covergece: so we expect Now let s differetiate agai: so f (x) = c x = c + 2c 2 x + 3c 3 x 2 + 4c 4 x 3 +, f () = c. f (x) = 2 ( )c x 2 = 2c 2 + 6c 3 x + 2c 4 x 2 +, f () = 2c 2. Differetiatig a third time (formally) ad settig x = gives The geeral rule appears to be so we should have f (3) () = 6c 3. f () () =!c, c = f () ().! This procedure really is valid, accordig to the followig theorem whose log proof is omitted. Theorem 5.9. Ay fuctio represeted by a power series i a ope iterval ( r, r) is ifiitely differetiable i ( r, r) ad its derivatives ca be computed by termwise differetiatio of the power series. This meas our previous calculatios are justified: if a fuctio f(x) ca be writte i the form c x i a iterval aroud the we must have (5.) c = f () ().! I particular, a fuctio has at most oe expressio as a power series c x aroud. Ad a fuctio which is ot ifiitely differetiable aroud will defiitely ot have a power series represetatio c x. For istace, x has o power series represetatio of this form sice it is ot differetiable at. Remark 5.. We ca also cosider power series f(x) = c (x a), whose iterval of covergece is cetered at a. I this case the coefficiets are give by the formula c = f () (a)/!. For simplicity we will focus o power series cetered at oly.
18 8 KEITH CONRAD Remark 5.. Eve if a power series coverges o a closed iterval [ r, r], the power series for its derivative eed ot coverge at the edpoits. Cosider f(x) = ( ) x 2 /. It coverges for x but the derivative series is 2( ) x 2, which coverges for x <. Remark 5.2. Because a power series ca be differetiated (iside its iterval of covergece) termwise, we ca discover solutios to a differetial equatio by isertig a power series with ukow coefficiets ito the differetial equatio to get relatios betwee the coefficiets. The a few iitial coefficiets, oce chose, should determie all the higher oes. After we compute the radius of covergece of this ew power series we will have foud a solutio to the differetial equatio i a specific iterval. This idea goes back to Newto. It does ot ecessarily provide us with all solutios to a differetial equatio, but it is oe of the stadard methods to fid some solutios. Example 5.3. If x < the x =. Differetiatig both sides, for x < we have x Multiplyig by x gives us x = x = ( x) 2. x ( x) 2, so we have fially foud a closed form expressio for a ifiite series we first met back i Example 3.9. Example 5.4. If there is a power series represetatio c x for e x the (5.) shows c = /!. The oly possible way to write e x as a power series is x! = + x + x2 2 + x3 3! + x4 4! +. Is this really a valid formula for e x? Well, we checked before that this power series coverges for all x. Moreover, calculatig the derivative of this series reproduces the series agai, so this series is a fuctio satisfyig y (x) = y(x). The oly solutios to this differetial equatio are Ce x ad we ca recover C by settig x =. Sice the power series has costat term (so its value at x = is ), its C is, so this power series is e x : e x = x!. Loosely speakig, this meas the polyomials, + x, + x + x2 2, + x + x2 2 + x3 6,... are good approximatios to e x. (Where they are good will deped o how far x is from.) I particular, settig x = gives a ifiite series represetatio of the umber e: (5.2) e =! = ! + 4! + Formula (5.2) ca be used to verify a iterestig fact about e. Theorem 5.5. The umber e is irratioal.
19 Proof. (Fourier) For ay, ( e = + 2! + 3! + + )! ( = + 2! + 3! + +! INFINITE SERIES 9 ( + ) +! ) ( + 2)! + ( + )! + ( + + ( + 2)( + ) + The secod term i paretheses is positive ad bouded above by the geometric series + + ( + ) 2 + ( + ) 3 + =. Therefore ( < e + 2! + 3! + + )!!. Write the sum + /2! + + /! as a fractio with commo deomiator!, say as p /!. Clear the deomiator! to get (5.3) <!e p. So far everythig we have doe ivolves o uproved assumptios. Now we itroduce the ratioality assumptio. If e is ratioal, the!e is a iteger whe is large (sice ay iteger is a factor by! for large ). But that makes!e p a iteger located i the ope iterval (, ), which is absurd. We have a cotradictio, so e is irratioal. We retur to the geeral task of represetig fuctios by their power series. Eve whe a fuctio is ifiitely differetiable for all x, its power series could have a fiite radius of covergece. Example 5.6. Viewig /( + x 2 ) as /( ( x 2 )) we have the geometric series formula + x 2 = ( x 2 ) = ( ) x 2 whe x 2 <, or equivaletly x <. The series has a fiite iterval of covergece (, ) eve though the fuctio /( + x 2 ) has o bad behavior at the edpoits: it is ifiitely differetiable at every real umber x. Example 5.7. Let f(x) = x /. This coverges for x <. For x < we have f (x) = x = x = x. Whe x <, a atiderivative of /( x) is log( x). Sice f(x) ad log( x) have the same value (zero) at x = they are equal: log( x) = x for x <. Replacig x with x ad egatig gives (5.4) log( + x) = ( ) for x <. The right side coverges at x = (alteratig series) ad diverges at x =. It seems plausible, sice the series equals log( + x) o (, ), that this should remai true at the x ).
20 2 KEITH CONRAD boudary: does ( ) / equal log 2? (Notice this is the alteratig harmoic series.) We will retur to this later. I practice, if we wat to show a ifiitely differetiable fuctio equals its associated power series (f () ()/!)x o some iterval aroud we eed some kid of error estimate for the differece betwee f(x) ad the polyomial N = (f () ()/!)x. Whe the estimate goes to as N we will have proved f(x) is equal to its power series. Theorem 5.8. If f(x) is ifiitely differetiable the for all N N f () () f(x) = x + R N (x),! where R N (x) = N! Proof. Whe N = the desired result says = x f(x) = f() + f (N+) (t)(x t) N dt. x f (t) dt. This is precisely the fudametal theorem of calculus! We obtai the N = case from this by itegratio by parts. Set u = f (t) ad dv = dt. The du = f (t) dt ad we (cleverly!) take v = t x (rather tha just t). The x f (t) dt = f t=x x (t)(t x) f (t)(t x) dt so = f ()x + f(x) = f() + f ()x + t= x f (t)(x t) dt, x f (t)(x t) dt. Now apply itegratio by parts to this ew itegral with u = f (t) ad dv = (x t) dt. The du = f (3) (t) dt ad (cleverly) use v = (/2)(x t) 2. The result is so x f (t)(x t) dt = f () x x x f (3) (t)(x t) 2 dt, f(x) = f() + f ()x + f () x 2 + f (3) (t)(x t) 2 dt. 2! 2 Itegratig by parts several more times gives the desired result. We call R N (x) a remaider term. Corollary 5.9. If f is ifiitely differetiable the the followig coditios at the umber x are equivalet: f(x) = (f () ()/!)x, R N (x) as N. Proof. The differece betwee f(x) ad the Nth partial sum of its power series is R N (x), so f(x) equals its power series precisely whe R N (x) as N.
21 INFINITE SERIES 2 Example 5.2. I (5.4) we showed log( + x) equals ( ) x / for x <. What about at x =? For this purpose we wat to show R N () as N whe f(x) = log( + x). To estimate R N () we eed to compute the higher derivatives of f(x): f (x) = + x, f (x) = ( + x) 2, f (3) 2 (x) = ( + x) 3, f (4) 6 (x) = ( + x) 4, ad i geeral for. Therefore so R N () = N! f () (x) = ( ) ( )! ( + x) ( ) N N! ( + t) N+ ( t)n dt = ( ) N ( t) N dt, ( + t) N+ R N () dt ( + t) N+ = N N2 N as N. Sice the remaider term teds to, log 2 = ( ) /. Example 5.2. Let f(x) = si x. Its power series is ( ) x2+ (2 + )! = x x3 3! + x5 5! x7 7! +, which coverges for all x by the ratio test. We will show si x equals its power series for all x. This is obvious if x =. Sice si x has derivatives ± cos x ad ± si x, which are bouded by, for x > R N (x) = x N! f (N+) (t)(x t) N dt N! N! x x = xn+ x t N dt x N dt. N! This teds to as N, so si x does equal its power series for x >. If x < the we ca similarly estimate R N (x) ad show it teds to as N, but we ca also use a little trick because we kow si x is a odd fuctio: if x < the x > so si x = si( x) = ( ) ( x)2+ (2 + )! from the power series represetatio of the sie fuctio at positive umbers. Sice ( x) 2+ = x 2+, si x = ( ) x2+ x2+ (2 + )! = ( ) (2 + )!,
22 22 KEITH CONRAD which shows si x equals its power series for x < too. Example By similar work we ca show cos x equals its power series represetatio everywhere: cos x = ( ) x2 (2)! = x2 2! + x4 4! x6 6! +. The power series for si x ad cos x look like the odd ad eve degree terms i the power series for e x. Is the expoetial fuctio related to the trigoometric fuctios? This idea is suggested because of their power series. Sice e x has a series without the ( ) factors, we ca get a match betwee these series by workig ot with e x but with e ix, where i is a square root of. We have ot defied ifiite series (or eve the expoetial fuctio) usig complex umbers. However, assumig we could make sese of this the we would expect to have e ix = (ix)! = + ix x2 2! ix3 = x2 2! + x4 4! + i = cos x + i si x. For example, settig x = π would say e iπ =. Replacig x with x i the formula for e ix gives 3! + x4 4! + ix5 5! + (x x3 3! + x5 5! e ix = cos x i si x. Addig ad subtractig the formulas for e ix ad e ix lets us express the real trigoometric fuctios si x ad cos x i terms of (complex) expoetial fuctios: cos x = eix + e ix, si x = eix e ix. 2 2i Evidetly a deeper study of ifiite series should make systematic use of complex umbers! The followig questio is atural: who eeds all the remaider estimate busiess from Theorem 5.8 ad Corollary 5.9? After all, why ot just compute the power series for f(x) ad fid its radius of covergece? That has to be where f(x) equals its power series. Alas, this is false. Example Let f(x) = { e /x2, if x,, if x =. It ca be show that f(x) is ifiitely differetiable at x = ad f () () = for all, so the power series for f(x) has every coefficiet equal to, which meas the power series is the zero fuctio. But f(x) if x, so f(x) is ot equal to its power series aywhere except at x =. Example 5.23 shows that if a fuctio f(x) is ifiitely differetiable for all x ad the power series (f () ()/!)x coverges for all x, f(x) eed ot equal its power series aywhere except at x = (where they must agree sice the costat term of the power series is f()). This is why there is otrivial cotet i sayig a fuctio ca be represeted by its power series o some iterval. The eed for remaider estimates i geeral is importat. )
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