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If the series ∞n=1 bn converges, then the sequence {Bn }, which is nondecreasing, tends to a limit B. But then An ≤ Bn ≤ B forall n ∈ N, and so the sequence An of partial sums of the series ∞n=1 an is bounded.By the criterion∞ for convergence of a series with nonnegative terms (Theorem 7),the series n=1 an converges.The second assertion of the theorem follows from what has just been provedthrough proof by contradiction.11< n12 < (n−1)nfor n ≥ 2, we conclude that the seriesExample 24 Since n(n+1)∞ 1∞1n=1 n2 andn=1 n(n+1) converge or diverge together.11But the latter series can be summed directly, by observing that k(k+1)= k1 − k+1,n∞111and therefore k=1 k(k+1) = 1 − n+1 . Hence n=1 n(n+1) = 1. Consequently the∞ 11π2series ∞n=1 n2 converges.
It is interesting thatn=1 n2 = 6 , as will be provedbelow.Example 25 It should be observed that the comparison theorem applies only to seand bn = 0, for example, weries with nonnegative terms.Indeed, if we set an = −n∞bconvergeswhilehave an < bn and the series ∞nn=1n=1 an diverges.Corollary8 (The Weierstrass M-test for absolute convergence) Let ∞n=1 an and∞n for alln=1 bn be series.
Suppose there exists an index N ∈ N such that |an | ≤ b∞n > N . Then a sufficientconditionforabsoluteconvergenceoftheseriesn=1 an∞is that the series n=1 bn converge.3.1 The Limit of a Sequence99Proof In fact, by the comparison theorem the series ∞n=1|an | will then converge,and that is what is meant by the absolute convergence of ∞n=1 an .This important sufficiency test for absolute convergence is often stated briefly asfollows: If the terms of a series are majorized (in absolute value) by the terms of aconvergent numerical series, then the original series converges absolutely.sin nsin nExample 26 The series ∞n=1 n2 converges absolutely, since | n2 | ≤∞ 1series n=1 n2 converges, as we saw in Example 24.1n2and the√nCorollary 9 (Cauchy’s test) Let ∞n=1 an be a given series and α = limn→∞ |an |.Then the following are true:a) if α < 1, the series ∞n=1 an converges absolutely;b) if α > 1, the series ∞n=1 an diverges;c) there exist both absolutely convergent and divergent series for which α = 1.Proof a) If α < 1, we can choose q ∈ R such that√α < q < 1.
Fixing q, by definitionof the superior limit, we find N ∈ N such that n |an | < q for all n > N . Thus wenshall have |an | < q n for n > N , and since the series ∞n=1 q converges for |q| < 1,it follows from the comparison theorem or from the Weierstrass criterion that theseries ∞n=1 an converges absolutely.√b) Since α is a partial limit of the sequence{ n |an |} (Proposition 1), there exists#a subsequence {ank } such that limn→∞ nk |ank | = α.
Hence if α > 1, there existsso the necessary condition for converK ∈ N such that |ank | > 1 for all k > K, anddiverges.gence (an → 0) does not hold for the series ∞n=1 an . It therefore∞ 11c) We already know that the series ∞divergesandn=1 nn=1 n2 converges1(absolutely, since | n12 | = n12 ). At the same time, limn→∞ n n1 = limn→∞ √nn = 111 2and limn→∞ n n12 = limn→∞ √n 2 = limn→∞ ( √n n ) = 1.nExample 27 Let us investigate the values of x ∈ R for which the series∞n2 + (−1)n x nn=1converges.√We compute α = limn→∞ n |(2 + (−1)n )n x n | = |x| limn→∞ |2 + (−1)n | = 3|x|.Thus for |x| < 13 the series converges and even absolutely, while for |x| > 3 theseries diverges.
The case |x| = 13 requires separate consideration. In the presentcase that is an elementary task, since for |x| = 13 and n even (n = 2k), we have(2 + (−1)2k )2k x 2k = 32k ( 13 )2k = 1. Therefore the series diverges, since it does notfulfill the necessary condition for convergence.1003LimitsCorollary 10 (d’Alembert’s6 test) Suppose the limit limn→∞ | an+1an | = α exists fora.Then,the series ∞n=1 na) if α < 1, the series ∞n=1 an converges absolutely;b) if α > 1, the series ∞n=1 an diverges;c) there exist both absolutely convergent and divergent series for which α = 1.Proof a) If α < 1, there exists a number q such that α < q < 1.
Fixing q and usingproperties of limits, we find an index N ∈ N such that | an+1an | < q for n > N . Since afinite number of terms has no effect on the convergence of a series, we shall assumewithout loss of generality that | an+1an | < q for all n ∈ N.Since an+1 an a2 an+1 ···· = a a , a ann−111nwe find that |an+1 | ≤ |a1 | · q n . But the series ∞n=1 |a1 |q converges (its sum is|a1 |q∞obviously 1−q ), so that the series n=1 an converges absolutely.| > 1, that is, |an | <b) If α > 1, then from some index N ∈ N on we have | αan+1n|an+1 |, and thecondition an → 0, which is necessary for convergence, does not holdfor the series ∞n=1 an .∞ 11c) As in the case of Cauchy’s test, the series ∞n=1 n andn=1 n2 provide examples.Example 28 Let us determine the values of x ∈ R for which the series∞1 nxn!n=1converges.For x = 0 it obviously converges absolutely.|x|For x = 0 we have limn→∞ | an+1an | = limn→∞ n+1 = 0.Thus, this series converges absolutely for every value of x ∈ R.Finally, let us consider another special, but frequently encountered class of series,namely those whose terms form a monotonic sequence.
For such series we have thefollowing necessary and sufficient condition:∞Proposition 2 (Cauchy)∞ kIf a1 ≥ a2 ≥ · · · ≥ 0, the series n=1 an converges if andonly if the series k=0 2 a2k = a1 + 2a2 + 4a4 + 8a8 + · · · converges.Proof Sincea2 ≤ a2 ≤ a1 ,6 J.L. d’Alembert (1717–1783) – French scholar specializing in mechanics. He was a member ofthe group of philosophers who wrote the Encyclopédie.3.1 The Limit of a Sequence1012a4 ≤ a3 + a4 ≤ 2a2 ,4a8 ≤ a5 + a6 + a7 + a8 ≤ 4a4 ,...2n a2n+1 ≤ a2n +1 + · · · + a2n+1 ≤ 2n a2n ,by adding these inequalities, we find1(Sn+1 − a1 ) ≤ A2n+1 − a1 ≤ Sn ,2where Ak = a1 + · · · + ak and Sn = a1 + 2a2 + · · · + 2n a2n are the partial sumsof the two series in question.
The sequences {Ak } and {Sn } are nondecreasing, andhence from these inequalities one can conclude that they are either both boundedabove or both unbounded above. Then, by the criterion for convergence of serieswith nonnegative terms, it follows that the two series indeed converge or divergetogether.This result implies a useful corollary.Corollary The series∞1n=1 npconverges for p > 1 and diverges for p ≤ 1.7Proof If p ≥ 0, the proposition implies that the series converges or diverges simultaneously with the series∞k=0∞2kk121−p ,=kp(2 )k=0and a necessary and sufficient condition for the convergence of this series is thatq = 21−p < 1, that is, p > 1.1If p ≤ 0, the divergence of the series ∞n=1 np is obvious, since all the terms ofthe series are not smaller than 1.The importance of this corollary is that the series ∞n=1comparison series to study the convergence of other series.1npis often used as ac.
The Number e as the Sum of a SeriesTo conclude our study of series we return once again to the number e and obtain aseries the provides a very convenient way of computing it.7 Up to now in this book the number np has been defined formally only for rational values of p, sothat for the moment the reader is entitled to take this proposition as applying only to values of pfor which np is defined.1023LimitsWe shall use Newton’s binomial formula to expand the expression (1 + n1 )n .Those who are unfamiliar with this formula from high school and have not solvedpart g) of Exercise 1 in Sect. 2.2 may omit the present appendix on the number ewith no loss of continuity and return to it after studying Taylor’s formula, of whichNewton’s binomial formula may be regarded as a special case.We know that e = limn→∞ (1 + n1 )n .By Newton’s binomial formula1+1nn=1+n 1 n(n − 1) 1++ ··· +1! n2!n2n(n − 1) · · · (n − k + 1) 11+ ··· + n =kk!nn11112=1+1+1−+ ··· +1−1−× ··· ×2!nk!nn 11n−1k−1+ ··· +1−··· 1 −.× 1−nn!nn+1Setting (1 + n1 )n = en and 1 + 1 + · · · 2!1 + · · · + n!= sn , we thus have en < sn(n = 1, 2, .
. .).On the other hand, for any fixed k and n ≥ k, as can be seen from the sameexpansion, we have 1111k−11+1+1−+ ··· +1−··· 1 −< en .2!nk!nnAs n → ∞ the left-hand side of this inequality tends to sk and the right-hand sideto e. We can now conclude that sk ≤ e for all k ∈ N.But then from the relationsen < sn ≤ ewe find that limn→∞ sn = e.In accordance with the definition of the sum of a series, we can now writee=1+111+ + ··· + + ··· .1! 2!n!This representation of the number e is very well adapted for computation.Let us estimate the difference e − sn :11++ ··· =(n + 1)! (n + 2)!$%111=1+++ ··· <(n + 1)!n + 2 (n + 2)(n + 3)0 < e − sn =3.1 The Limit of a Sequence103<%$111++···=1+(n + 1)!n + 2 (n + 2)2=11n+21.=<1(n + 1)! 1 − n+2n!(n + 1)2 n!nThus, in order to make the absolute error in the approximation of e by sn less11< 1000.
This condition is already satisfied by s6 .than, say 10−3 , it suffices that n!nLet us write out the first few decimal digits of e:e = 2.7182818284590 . . . .This estimate of the difference e − sn can be written as the equalitye = sn +θn,n!nwhere 0 < θn < 1.It follows immediately from this representation of e that it is irrational.
Indeed,if we assume that e = pq , where p, q ∈ N, then the number q!e must be an integer,whileθqq! q!q! θqq!e = q! sq += q! + + + · · · + + ,q!q1! 2!q!qθand then the number qq would have to be an integer, which is impossible.For the reader’s information we note that e is not only irrational, but also transcendental.3.1.5 Problems and Exercises1. Show that a number x ∈ R is rational if and only if its q-ary expression in anybase q is periodic, that is, from some rank on it consists of periodically repeatingdigits.2.
A ball that has fallen from height h bounces to height qh, where q is a constant coefficient 0 < q < 1. Find the time that elapses until it comes to rest and thedistance it travels through the air during that time.3. We mark all the points on a circle obtained from a fixed point by rotations of thecircle through angles of n radians, where n ∈ Z ranges over all integers. Describeall the limit points of the set so constructed.4. The expressionn1 +1n2 +n3 +1...1nk−1 + n1k1043Limitswhere nk ∈ N, is called a finite continued fraction, and the expressionn1 +1n2 +1n3 +...is called an infinite continued fraction.
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