Math 141: Lecture 19

Math 141: Lecture 19 Convergence of infinite series Bob Hough November 16, 2016 Bob Hough Math 141: Lecture 19 November 16, 2016 1 / 44

Series of positive terms Recall that, given a sequence {a n } n=1, we say that its series n=1 a n converges if the sequence of partial sums {s n } n=1 s n = n a k, converges. We begin by giving some criteria for the convergence in the case when the terms a n are non-negative. k=1 Bob Hough Math 141: Lecture 19 November 16, 2016 2 / 44

Comparison test Theorem (Comparison test) Let {a n } n=1 and {b n} n=1 be two sequences such that a n 0 and b n 0 for all n 1. Suppose there is a number N and a constant c > 0 such that a n cb n for all n > N. Then convergence of n=1 b n implies convergence of n=1 a n. In this case, we say that the sequence b n dominates the sequence a n. Bob Hough Math 141: Lecture 19 November 16, 2016 3 / 44

Comparison test Proof of the comparison test. Suppose that n=1 b n converges. Given ɛ > 0, apply the Cauchy property of the sequence of partial sums of b n to choose M > N such that m > n > M implies It follows that n k=m+1 n k=m+1 a n b n < ɛ c. n k=m+1 cb n < ɛ, so the sequence of partial sums of a n is Cauchy, hence converges. Bob Hough Math 141: Lecture 19 November 16, 2016 4 / 44

Limit comparison test Theorem (Limit comparison test) Let {a n } n=1 and {b n} n=1 be sequences satisfying a n > 0 and b n > 0 for all n 1. Suppose that a n lim = 1. n b n Then n=1 a n converges if and only if n=1 b n converges. Proof. The condition implies that there is N such that n > N implies a n 2 b n 2a n. It now follows from the comparison test that n=1 a n converges if and only if n=1 b n converges. Bob Hough Math 141: Lecture 19 November 16, 2016 5 / 44

Examples Recall that last class we checked n=1 1 n 2 + n = n=1 1 n(n + 1) = n=1 ( 1 n 1 ) = 1 n + 1 n since the series telescopes. Since lim 2 +n n = lim n 2 n 1 + 1 n = 1, it follows that n=1 1 converges. n 2 By comparison, it follows that n=1 1 n converges for all s 2. s Bob Hough Math 141: Lecture 19 November 16, 2016 6 / 44

Examples Recall that the harmonic series satisfies check that by noting that and sin 1 n lim n 1 n n=1 lim n 1 n(n + 10) =, n(n + 10) n = lim n n=1 1 n =. One may sin 1 n = n=1 1 + 10 n = 1, ( ( )) ( ) 1 1 1 = lim n n n + O n 3 = lim 1 + O n n 2 = 1. Bob Hough Math 141: Lecture 19 November 16, 2016 7 / 44

Integral test Theorem (Integral test) Let f be a positive decreasing function, defined for all real x 1. The infinite series n=1 f (n) converges if and only if the improper integral 1 f (x)dx converges. Bob Hough Math 141: Lecture 19 November 16, 2016 8 / 44

Integral test Proof of the integral test. We first check that the improper integral converges if and only if the sequence { t n = n 1 f (x)dx} n=1 converges. Recall that convergence of the improper integral is defined by the x existence of the limit lim x 1 f (t)dt. Evidently convergence of the integral entails convergence of the sequence as a special case. To check the reverse direction, note that for n x n + 1, t n x 1 f (t)dt t n+1, and hence, if t n L < ɛ for all n N, then x f (t)dt L < ɛ for all x N also. 1 Bob Hough Math 141: Lecture 19 November 16, 2016 9 / 44

Integral test Proof of the integral test. We now show that {t n } n=1 converges if and only if the sequence of partial sums {s n = n k=1 f (k)} k=1 converges. Given m > n, note that by taking upper and lower step functions for the integral, m 1 k=n f (k) t m t n = m n f (t)dt m k=n+1 f (k) and hence {t n } n=1 is Cauchy if and only if {s n} n=1 proves the claim. is Cauchy, which Bob Hough Math 141: Lecture 19 November 16, 2016 10 / 44

Examples We can show that the series n=1 1 n, s real, converges if and only if s s > 1. To check this, note that t n = n 1 dt t s = { n 1 s 1 1 s s 1 log n s = 1. This sequence converges if and only if s > 1. 1 For complex s = σ + it, n = 1 s n exp (it log n). Since σ exp (it log n) = 1, the sum ζ(s) = n=1 converges absolutely if σ = R(s) > 1. This function is called the Riemann zeta function. 1 n s Bob Hough Math 141: Lecture 19 November 16, 2016 11 / 44

Examples The study of the analytic properties of the Riemann zeta function and related functions is responsible for a great deal of the information that we know about the prime numbers. Bob Hough Math 141: Lecture 19 November 16, 2016 12 / 44

Root test Theorem (Root test) Let n=1 a n be a series of non-negative terms. Suppose that a 1 n n R, n. If R < 1 then the series converges. If R > 1 then the series diverges. Bob Hough Math 141: Lecture 19 November 16, 2016 13 / 44

Root test Proof of the root test. Let x = R+1 2 so that R < x < 1 if x < 1 and 1 < x < R if R > 1. Suppose R < 1. Then, for all n sufficiently large, a n < x n. The series converges by comparison with the geometric series n=1 x n. Suppose instead that R > 1. Then for all n sufficiently large, a n > x n so that the series diverges by comparison with the geometric series n=1 x n. Bob Hough Math 141: Lecture 19 November 16, 2016 14 / 44

Example We apply the root test to the series ( ) n 2 n n=1 n+1. Calculate a 1 n n = Thus the series converges. ( ) ( n n 1 = n + 1 1 + 1 n ) n 1 e. Bob Hough Math 141: Lecture 19 November 16, 2016 15 / 44

Ratio test Theorem Let n=1 a n be a sum of positive terms such that a n+1 a n L, n. The series converges if L < 1 and diverges if L > 1. Bob Hough Math 141: Lecture 19 November 16, 2016 16 / 44

Ratio test Proof of the ratio test. If L > 1 then the series does not converge since the terms do not tend to 0. If L < 1, let L < x = L + 1 < 1. 2 Choose N sufficiently large, so that if n N then a n+1 a n < x. It follows that, for all n > N, n 1 a n = a N k=n a k+1 a k a N x n N = a N x N x n. Since a N x N is just a constant, the series converges by comparison with the geometric series n=1 x n. Bob Hough Math 141: Lecture 19 November 16, 2016 17 / 44

Examples Consider the series n=1 n! n. Applying the ratio test, as n, n a n+1 a n = Hence the series converges. ( ) (n + 1)! nn n n (n + 1) n+1 n! = = n + 1 1 (1 + 1/n) n 1 e. Bob Hough Math 141: Lecture 19 November 16, 2016 18 / 44

Edge cases The ratio and root tests are not useful in the case a n+1 a n 1 as n. Note that n=1 1 n = while n=1 1 converges. The ratios of n 2 consecutive terms in these cases are given by n n + 1 = 1 1 ( ) 1 n + O n 2 n 2, (n + 1) 2 = 1 2 ( ) 1 n + O n 2. The following two tests of Raabe and Gauss sometimes help to decide convergence in cases where the limiting ratio is 1. Bob Hough Math 141: Lecture 19 November 16, 2016 19 / 44

Raabe s test Theorem (Raabe s test) Let n=1 a n be a sum of positive terms. If there is an r > 0 and N 1 such that, for all n N, a n+1 a n 1 1 n r n then n=1 a n converges. If, for all n N, a n+1 a n 1 1 n then n=1 a n =. Bob Hough Math 141: Lecture 19 November 16, 2016 20 / 44

An inequality for the logarithm In the first part of the proof of Raabe s test we use the inequality log(1 + x) x, 1 < x. To check this inequality, note equality holds at x = 0. Also, x log(1 + x) has derivative 1 1 1+x, which changes sign at x = 0. It follows that the function decreases to 0, and increases thereafter. Bob Hough Math 141: Lecture 19 November 16, 2016 21 / 44

Raabe s test Proof of Raabe s test. Assume first that a n+1 a n 1 1+r n for n > N. Thus, for n > N, n 1 a n a N k=n ( 1 r + 1 ) a N exp k ( n 1 k=n There is some constant C (which depends on N) such that n 1 k=n 1 k C + log n. ) r + 1. k Hence, there is another constant c, depending on N, such that a n for n > N. The series thus converges by comparison with 1 n=1 n 1+r. c n 1+r Bob Hough Math 141: Lecture 19 November 16, 2016 22 / 44

Raabe s test Proof of Raabe s test. Now assume that a n+1 a n 1 1 n for n > N 1. Thus, for n > N, n 1 a n a N k=n ( 1 1 ). k Note that the product is equal to n 1 k=n k 1 k = N 1 n 1, since it telescopes. Since there is a constant c (depending on N) such that a n > c n 1 for all n > N, the series diverges by comparison with the harmonic series. Bob Hough Math 141: Lecture 19 November 16, 2016 23 / 44

Gauss s test Theorem (Gauss s test) Let n=1 a n be a series of positive terms. If there is an N 1, an s > 1 and an M > 0 such that for all n > N, a n+1 a n = 1 A n + f (n) n s where f (n) M for all n, then n=1 a n converges if A > 1 and diverges if A 1. Bob Hough Math 141: Lecture 19 November 16, 2016 24 / 44

Gauss s test Proof of Gauss s test. Let N be sufficiently large, and write, for n > N, n 1 a n = a N k=n a k+1 a k = a N n 1 k=n Since ( A k f (k) k tends to 0 as k we may write s ) log 1 A k + f (k) k = A s k + f (k) k + O ( ) 1 s k. Hence 2 a n = a N exp ( n 1 k=n ( A k + f (k) k s = a N exp ( A log n + O(1)). ( 1 A k + f (k) ) k s. + O ( )) ) 1 k 2 The series thus converges or diverges according as A > 1 or otherwise, by comparison with the sum n 1 n A. Bob Hough Math 141: Lecture 19 November 16, 2016 25 / 44

Examples Consider the series ( ) 1 3 5 (2n 1) k. n=1 2 4 6 (2n) The ratio a n a n 1 sequence is given by ( 2n 1 2n ) k ( = 1 1 2n ) k ( ( = exp k 1 2n + O = 1 k ( ) 1 2n + O n 2 ( ))) 1 n 2 of this as n. Hence the series converges for k > 2 and diverges for k 2 by Gauss s test. Bob Hough Math 141: Lecture 19 November 16, 2016 26 / 44

Power series Definition Let {a n } n=0. The series P(x) = n=0 a nx n is called the power series associated to {a n } n=0. The series n=0 a nx n is sometimes also called a (linear) generating function of {a n } n=0. We ll return to study power series in more detail in the next several lectures, but point out some basic properties now. Bob Hough Math 141: Lecture 19 November 16, 2016 27 / 44

Power series Theorem Let {a n } n=0 be a sequence. Define R = 0 if { a n 1 n } n=1 R = if a n 1 n 0, and is unbounded, 1 R = lim sup a n 1 n n otherwise. The power series P(x) = n=0 a nx n converges absolutely in the disc D R (0) = {x C : x < R} and diverges for x > R. Definition As defined, R is called the radius of convergence of the power series P(x) = n=0 a nx n. Bob Hough Math 141: Lecture 19 November 16, 2016 28 / 44

Power series Proof. First suppose R = 0, so that a n 1 n is unbounded. It follows that for any x 0, a n x n 1 n = a n 1 n x is unbounded. The power series P(x) diverges since its terms do not tend to 0. Next suppose R = so that a n 1 n 0. Then for any x C, a n x n 1 n = a n 1 n x 0, so that the power series converges absolutely, by comparison with a geometric series. Finally, suppose that R > 0 is finite. For x < R, lim sup n a n x n 1 n = lim sup a n 1 x n x = n R < 1. The series thus converges absolutely by comparison with a geometric series. If instead x > R, then the limsup is x R > 1, and the power series does not converge since its terms do not tend to 0. Bob Hough Math 141: Lecture 19 November 16, 2016 29 / 44

Solving linear recurrences A linear recurrence sequence is a sequence in which successive terms are defined as a linear combination of a bounded number of terms preceeding them (and possibly an auxiliary function). Examples include The Fibonacci numbers, F 0 = F 1 = 1, for n 2, F n = F n 1 + F n 2. The number of steps needed to solve the Towers of Hanoi with n discs, T 0 = 0, T n+1 = 2T n + 1. (T n = 2 n 1). The number of regions into which the plane is split by n lines in general position (pairwise non-parallel), L 0 = 1, L n+1 = L n + n + 1. (L n = 1 + ( ) n+1 2 ). Bob Hough Math 141: Lecture 19 November 16, 2016 30 / 44

Solving linear recurrences Although there are other methods, often the easiest way of solving linear recurrences is through writing down the power series generating function. Fibonacci numbers: Let f (x) = n=0 F nx n. By the recurrence relation, f (x) = 1 + x + = 1 + x + F n x n n=2 (F n 2 + F n 1 )x n n=2 = 1 + (x + x 2 )f (x). Hence f (x)(1 x x 2 ) = 1, or f (x) = 1 1 x x 2. Bob Hough Math 141: Lecture 19 November 16, 2016 31 / 44

Solving linear recurrences Recall f (x) = n=0 F nx n = 1 1 x x 2. Let φ = 1+ 5 2, 1 φ = 5 1 2 so that (1 x x 2 ) = (1 φx)(1 + x φ ). Write in partial fractions, 1 1 x x 2 = A 1 φx + B 1 + x φ with A + B = 1, A φ Bφ = 0 so that A = φ2 Matching coefficients of x n, F n = φ2 1 + φ 2 φn + ( 1)n 1 + φ 2 φ n., B = 1. 1+φ 2 1+φ 2 The radius of convergence of the power series f (x) is 1 φ. Bob Hough Math 141: Lecture 19 November 16, 2016 32 / 44

Partial summation Theorem (Abel summation) Let {a n } n=1 and {b n} n=1 be two sequences of complex numbers, and let n A n = a k. k=1 Then n n a k b k = A n b n+1 + A k (b k b k+1 ). k=1 k=1 Bob Hough Math 141: Lecture 19 November 16, 2016 33 / 44

Partial summation Proof. Define A 0 = 0. Thus, since a k = A k A k 1, k = 1, 2,..., one has n a k b k = k=1 = n (A k A k 1 )b k k=1 n A k b k k=1 = A n b n+1 + n A k b k+1 + A n b n+1 k=1 n A k (b k b k+1 ). k=1 Bob Hough Math 141: Lecture 19 November 16, 2016 34 / 44

Dirichlet s test Theorem (Dirichlet s test) Let {a n } be a sequence of complex terms, whose partial sums form a bounded sequence. Let {b n } be a decreasing sequence of positive real terms, which tends to 0. Then the series converges. a n b n n=1 This generalizes the alternating series test from last lecture. Bob Hough Math 141: Lecture 19 November 16, 2016 35 / 44

Dirichlet s test Proof of Dirichlet s test. Denote the sequence of partial sums of {a n } n=1 by {A n} n=1. Let M > 0 be such that A n M for all n. By Abel summation, the sequence of partial sums {s n = n k=1 a kb k } n=1 satisfy s n = A n b n+1 + n A k (b k b k+1 ). k=1 Note that A n b n+1 0 as n, while k=1 A k(b k b k+1 ) converges absolutely, since A k (b k b k+1 ) M (b k b k+1 ) = Mb 1. k=1 This proves the convergence. k=1 Bob Hough Math 141: Lecture 19 November 16, 2016 36 / 44

Examples Let θ R. The sum diverges otherwise. k=1 e2πikθ k converges if θ is not an integer and To see this, for θ Z, e 2πikθ = 1, so this is the harmonic series, which diverges. Otherwise, set z = e 2πiθ so z = 1, z 1. Then A n = n k=1 e2πikθ = z zn 1 z satisfies A n 2 1 z. Hence the convergence follows from Dirichlet s test. Bob Hough Math 141: Lecture 19 November 16, 2016 37 / 44

Abel s test Theorem (Abel s test) Let n=1 a n be a convergent complex series, and let {b n } n=1 be a monotonic, convergent sequence of real terms. Then converges. n=1 a n b n Bob Hough Math 141: Lecture 19 November 16, 2016 38 / 44

Abel s test Proof of Abel s test. As in the proof of Dirichlet s test, write A n = n k=1 a k, and assume that A n M. Let b = lim n b n. By Abel summation, n a k b k = A n b n+1 + k=1 n A k (b k b k+1 ). k=1 As n, lim n A n b n+1 = b k=1 a k converges. Also, A k (b k b k+1 ) M b k b k+1 = M b 1 b k=1 k=1 since the series on the right telescopes. Thus k=1 A k(b k b k+1 ) converges absolutely. Bob Hough Math 141: Lecture 19 November 16, 2016 39 / 44

Products of series Definition Let n=0 a n and n=0 b n be two series. Their Cauchy product is the series c n, c n = a 0 b n + a 1 b n 1 + + a n b 0. n=0 Bob Hough Math 141: Lecture 19 November 16, 2016 40 / 44

Example The following example shows that it is possible that n=1 a n and n=1 b n converge, but their Cauchy product n=1 c n does not converge. ( 1) n Let a n = b n = ( 1)n n+1. The series n=0 n+1 converges by the alternating series test. The Cauchy product has c n = ( 1) n n 1 k=0 k+1 n k+1. By the Inequality of the Arithmetic Mean-Geometric Mean, (k + 1)(n k + 1) n + 2 2, and hence c n n 2 k=0 n+2 = 2(n+1) n+2. Since c n 1 as n, the series does not converge. Bob Hough Math 141: Lecture 19 November 16, 2016 41 / 44

Products of series Theorem Let n=0 a n = A converge absolutely, and n=0 b n = B converge. Denote by {c n } n=0 the Cauchy product of {a n} n=0 and {b n} n=0. Then c n = AB. n=0 Bob Hough Math 141: Lecture 19 November 16, 2016 42 / 44

Products of series Proof of Cauchy Product formula. See Rudin, pp. 74-75. Define partial sums A n = n k=0 a n, B n = n k=0 b k, C n = n k=0 c k. Set β n = B n B. Write C n = a 0 b 0 + (a 0 b 1 + a 1 b 0 ) + + (a 0 b n + a 1 b n 1 + + a n b 0 ) = a 0 B n + a 1 B n 1 + + a n B 0 = a 0 (B + β n ) + a 1 (B + β n 1 ) + a n (B + β 0 ) = A n B + a 0 β n + a 1 β n 1 + + a n β 0. Define γ n = a 0 β n + a 1 β n 1 + + a n β 0. Since A n B AB as n, it suffices to check that γ n 0. Bob Hough Math 141: Lecture 19 November 16, 2016 43 / 44

Products of series Proof of Cauchy Product formula. See Rudin, pp. 74-75. Recall that β n = B n B tends to 0 with n, and that we wish to show that γ n = a 0 β n + a 1 β n 1 + + a n β 0 tends to 0 also. Define α = n=0 a n. Given ɛ > 0, choose N such that n > N implies β n < ɛ. Bound γ n N a n k β k + k=0 n k=n+1 a n k β k N a n k β k + ɛα. Since a n 0 as n, if n is sufficientlly large, then N k=0 a n kβ k < ɛ, whence γ n (1 + α)ɛ. Letting ɛ 0 completes the proof. k=0 Bob Hough Math 141: Lecture 19 November 16, 2016 44 / 44