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Chapter 4

Series

Divergent series are the devil, and it is a shame to base on them any demonstration whatsoever. (Niels Henrik Abel, 1826) This series is divergent, therefore we may be able to do something with it. (Oliver Heaviside, quoted by Kline) In this chapter, we apply our results for sequences to series, or innite sums. The convergence and sum of an innite series is dened in terms of its sequence of nite partial sums.

4.1. Convergence of series

A nite sum of real numbers is well-dened by the algebraic properties ofR, but in order to make sense of an innite series, we need to consider its convergence. We say that a series converges if its sequence of partial sums converges, and in that case we dene the sum of the series to be the limit of its partial sums. Denition 4.1.Let (an) be a sequence of real numbers. The series 1 X n=1a n converges to a sumS2Rif the sequence (Sn) of partial sums S n=nX k=1a k converges toSasn! 1. Otherwise, the series diverges.

If a series converges toS, we write

S=1X n=1a n:59

604. SeriesWe also say a series diverges to1if its sequence of partial sums does. As for

sequences, we may start a series at other values ofnthann= 1 without changing its convergence properties. It is sometimes convenient to omit the limits on a series when they aren't important, and write it asPan. Example 4.2.Ifjaj<1, then the geometric series with ratioaconverges and its sum is1X n=0a n=11a: This series is simple enough that we can compute its partial sums explicitly, S n=nX k=0a k=1an+11a: As shown in Proposition 3.31, ifjaj<1, thenan!0 asn! 1, so thatSn!

1=(1a), which proves the result.

The geometric series diverges to1ifa1, and diverges in an oscillatory fashion ifa 1. The following examples consider the casesa=1 in more detail.

Example 4.3.The series

1X n=11 = 1 + 1 + 1 +::: diverges to1, since itsnth partial sum isSn=n.

Example 4.4.The series

1X n=1(1)n+1= 11 + 11 +::: diverges, since its partial sums S n=(

1 ifnis odd;

0 ifnis even;

oscillate between 0 and 1. This series illustrates the dangers of blindly applying algebraic rules for nite sums to series. For example, one might argue that

S= (11) + (11) + (11) += 0 + 0 + 0 += 0;

or that

S= 1 + (1 + 1) + (1 + 1) += 1 + 0 + 0 += 1;

or that

1S= 1(11 + 11 +:::) = 11 + 11 + 1 =S;

so 2S= 1 orS= 1=2. The Italian mathematician and priest Luigi Grandi (1710) suggested that these results were evidence in favor of the existence of God, since they showed that it was possible to create something out of nothing.

4.1. Convergence of series61Telescoping series of the form

1X n=1(anan+1) are another class of series whose partial sums S n=a1an+1 can be computed explicitly and then used to study their convergence. We give one example.

Example 4.5.The series

1X n=11n(n+ 1)=112+123+134+145+::: converges to 1. To show this, we observe that

1n(n+ 1)=1n

1n+ 1;

so nX k=11k(k+ 1)=nX k=1 1k 1k+ 1 11 12 +12 13 +13 14 ++1n 1n+ 1 = 11n+ 1; and it follows that 1X k=11k(k+ 1)= 1: A condition for the convergence of series with positive terms follows immedi- ately from the condition for the convergence of monotone sequences. Proposition 4.6.A seriesPanwith positive termsan0 converges if and only if its partial sums nX k=1a kM are bounded from above, otherwise it diverges to1.

Proof.The partial sumsSn=Pn

k=1akof such a series form a monotone increasing sequence, and the result follows immediately from Theorem 3.29 Although we have only dened sums of convergent series, divergent series are not necessarily meaningless. For example, the Cesaro sumCof a seriesPanis dened by

C= limn!11n

n X k=1S n; Sn=a1+a2++an:

624. SeriesThat is, we average the rstnpartial sums the series, and letn! 1. One can

prove that if a series converges toS, then its Cesaro sum exists and is equal toS, but a series may be Cesaro summable even if it is divergent. Example 4.7.For the seriesP(1)n+1in Example 4.4, we nd that 1n n X k=1S k=(

1=2 + 1=(2n) ifnis odd;

1=2 ifnis even;

since theSn's alternate between 0 and 1. It follows the Cesaro sum of the series is C= 1=2. This is, in fact, what Grandi believed to be the \true" sum of the series. Cesaro summation is important in the theory of Fourier series. There are also many other ways to sum a divergent series or assign a meaning to it (for example, as an asymptotic series), but we won't discuss them further here.

4.2. The Cauchy condition

The following Cauchy condition for the convergence of series is an immediate con- sequence of the Cauchy condition for the sequence of partial sums.

Theorem 4.8(Cauchy condition).The series

1X n=1a n converges if and only for every >0 there existsN2Nsuch thatn X k=m+1a k =jam+1+am+2++anj< for alln > m > N: Proof.The series converges if and only if the sequence (Sn) of partial sums is Cauchy, meaning that for every >0 there existsNsuch that jSnSmj= n X k=m+1a k < for alln > m > N; which proves the result. A special case of this theorem is a necessary condition for the convergence of a series, namely that its terms approach zero. This condition is the rst thing to check when considering whether or not a given series converges.

Theorem 4.9.If the series1X

n=1a n converges, then limn!1an= 0: Proof.If the series converges, then it is Cauchy. Takingm=n1 in the Cauchy condition in Theorem 4.8, we nd that for every >0 there existsN2Nsuch that janj< for alln > N, which proves thatan!0 asn! 1.

4.2. The Cauchy condition63Example 4.10.The geometric seriesPanconverges ifjaj<1 and in that case

a n!0 asn! 1. Ifjaj 1, thenan6!0 asn! 1, which implies that the series diverges. The condition that the terms of a series approach zero is not, however, sucient to imply convergence. The following series is a fundamental example.

Example 4.11.The harmonic series

1X n=11n = 1 +12 +13 +14 diverges, even though 1=n!0 asn! 1. To see this, we collect the terms in successive groups of powers of two, 1X n=11n = 1 +12 +13 +14 +15 +16 +17 +18 +19 +110
++116 >1 +12 +14 +14 +18 +18 +18 +18 +116
+116
++116 >1 +12 +12 +12 +12

In general, for everyn1, we have

2 n+1X k=11k = 1 +12 +nX j=12 j+1X k=2j+11k >1 +12 +nX j=12 j+1X k=2j+112 j+1 >1 +12 +nX j=112 n2 +32
so the series diverges. We can similarly obtain an upper bound for the partial sums, 2 n+1X k=11k <1 +12 +nX j=12 j+1X k=2j+112 j< n+32 These inequalities are rather crude, but they show that the series diverges at a logarithmic rate, since the sum of 2 nterms is of the ordern. This rate of divergence is very slow. It takes 12367 terms for the partial sums of harmonic series to exceed

10, and more than 1:51043terms for the partial sums to exceed 100.

A more rened argument, using integration, shows that lim n!1" nX k=11k logn# where

0:5772 is the Euler constant. (See Example 12.45.)

644. Series4.3. Absolutely convergent series

There is an important distinction between absolutely and conditionally convergent series.

Denition 4.12.The series1X

n=1a n converges absolutely if 1X n=1janjconverges; and converges conditionally if 1X n=1a nconverges, but1X n=1janjdiverges: We will show in Proposition 4.17 below that every absolutely convergent series converges. For series with positive terms, there is no dierence between convergence and absolute convergence. Also note from Proposition 4.6 thatPanconverges absolutely if and only if the partial sumsPn k=1jakjare bounded from above. Example 4.13.The geometric seriesPanis absolutely convergent ifjaj<1.

Example 4.14.The alternating harmonic series,

1X n=1(1)n+1n = 112 +13 14 is not absolutely convergent since, as shown in Example 4.11, the harmonic series diverges. It follows from Theorem 4.30 below that the alternating harmonic series converges, so it is a conditionally convergent series. Its convergence is made possible by the cancelation between terms of opposite signs. As we show next, the convergence of an absolutely convergent series follows from the Cauchy condition. Moreover, the series of positive and negative terms in an absolutely convergent series converge separately. First, we introduce some convenient notation. Denition 4.15.The positive and negative parts of a real numbera2Rare given by a aifa >0;

0 ifa0;a=(

0 ifa0;

jajifa <0:

It follows, in particular, that

0a+;a jaj; a=a+a;jaj=a++a:

We may then split a series of real numbers into its positive and negative parts. Example 4.16.Consider the alternating harmonic series 1X n=1a n= 112 +13 14 +15 16

4.3. Absolutely convergent series65Its positive and negative parts are given by

1 X n=1a +n= 1 + 0 +13 + 0 +15 + 0 +:::; 1 X n=1a n= 0 +12 + 0 +14 + 0 +16 Both of these series diverge to innity, since the harmonic series diverges and 1 X n=1a +n>1X n=1a n=12 1 X n=11n Proposition 4.17.An absolutely convergent series converges. Moreover, 1 X n=1a n converges absolutely if and only if the series 1 X n=1a +n;1X n=1a n of positive and negative terms both converge. Furthermore, in that case 1 X n=1a n=1X n=1a +n1X n=1a n;1X n=1janj=1X n=1a +n+1X n=1a n: Proof.IfPanis absolutely convergent, thenPjanjis convergent, so it satises the Cauchy condition. Since n X k=m+1a k nX k=m+1jakj; the series Panalso satises the Cauchy condition, and therefore it converges.

For the second part, note that

0nX k=m+1jakj=nX k=m+1a k+nX k=m+1a k; 0nX k=m+1a knX k=m+1jakj; 0nX k=m+1a knX k=m+1jakj; which shows that Pjanjis Cauchy if and only if bothPa+n,Panare Cauchy. It follows thatPjanjconverges if and only if bothPa+n,Panconverge. In that

664. Seriescase, we have

1X n=1a n= limn!1n X k=1a k = lim n!1 nX k=1a knX k=1a k! = lim n!1n X k=1a klimn!1n X k=1a k 1X n=1a +n1X n=1a n; and similarly for

Pjanj, which proves the proposition.

It is worth noting that this result depends crucially on the completeness ofR. Example 4.18.Suppose thata+n;an2Q+are positive rational numbers such that 1X n=1a +n=p2;1X n=1a n= 2p2; and letan=a+nan. Then 1Xquotesdbs_dbs29.pdfusesText_35

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