Homework 4 Solutions Math 171, Spring 2010 Please send corrections to [email protected] P1 P1 P1 p 26.5. Let n=1 an and n=1 bn be absolutely convergent series. Prove that the series n=1 janbnj converges. P1 P1 P1 Solution. Since n=1 janj and n=1 jbnj converge, by Theorem 23.1, n=1 janj + jbnj converges. Note that for all n, we have p p p janbnj = janjjbnj ≤ (janj + jbnj)(janj + jbnj) = janj + jbnj: P1 So by Theorem 26.3 (Comparison Test), since n=1 janj + jbnj converges absolutely, we see that P1 p n=1 janbnj converges absolutely, and hence converges as all terms are positive. 27.1. Find the radius of convergence R of each of the power series. Discuss the convergence of the power series at the points jx − tj = R. Solution. P1 x2n−1 (a) n=1 (2n−1)! Note that x2(n+1)−1 x2 lim (2(n+1)−1)! = lim = 0 n!1 x2n−1 n!1 (2n + 1)(2n) (2n−1)! so by the ratio test, the series converges absolutely for all x, and so R = 1. P1 n(x−1)n (b) n=1 2n Note that (n+1)(x−1)(n+1) (n+1) n + 1 x − 1 x − 1 lim 2 = lim = n!1 n(x−1)n n!1 n 2 2 2n so by the ratio test, the series converges absolutely if jx − 1j < 2 and diverges if jx − 1j > 2. Thus P1 P1 n R = 2. If jx − 1j = 2 then the power series diverges, since n=1 n and n=1(−1) n diverge. P1 n!(x+2)n (c) n=1 2n Note that (n+1)!(x+2)(n+1) ( (n+1) (n + 1)(x + 2) 0 if x = −2 lim 2 = lim = n!1 n!(x+2)n n!1 2 1 otherwise 2n so by the ratio test, the series diverges if jx + 2j > 0. Thus R = 0. If jx + 2j = 0 then the power series converges by the ratio test, or by the comment after Definition 27.1. P1 1=n n (d) n=1(n − 1)x By Theorems 16.7 and 25.1 (Alternating Series Test), the series converges when x = −1. 1 n By equality (16.3) in the proof of Theorem 16.7, we know (1 + n ) ≤ n. Taking n-th roots of 1 1=n P1 1 both sides and subtracting 1, we get n ≤ n − 1. We know that the series n=1 n diverges, hence P1 1=n so does n=1 n − 1. Thus the series diverges when x = 1. 1 Hence by Theorem 27.2, we have R = 1, and we have already discussed the behavior of the power series when jxj = 1. P1 (x−π)n (e) n=2 n(n−1) Note that (x−π)(n+1) n − 1 lim (n+1)n = lim (x − π) = jx − πj n!1 (x−π)n n!1 n + 1 n(n−1) so by the ratio test, the series converges absolutely if jx − πj < 1 and diverges if jx − πj > 1. Thus P1 1 P1 (−1)n R = 1. If jx−πj = 1 then the power series converges, since n=1 n(n−1) and n=1 n(n−1) converge. P1 n 27.2. Suppose that the power series n=0 an(x − t) has radius of convergence R. Let p be an integer. P1 p n Prove that the power series n=0 n an(x − t) has radius of convergence R. Solution. We consider thhree cases. 1=n 1 First case: 0 < R < 1. By Theorem 27.2, we have lim supn!1 janj = R . Note that p=n 1=n p p limn!1 n = (limn!1 n ) = 1 = 1 by Exercise 17.4 and Theorem 16.7. Note also that 1=n 1=n 1 fjanj g is a bounded sequence as lim supn!1 janj = R < 1. Hence Theorem 20.8 applies, and we get p 1=n p=n 1=n p=n 1=n 1 lim sup(n janj) = lim sup n janj = lim n lim sup janj = n!1 n!1 n!1 n!1 R P1 p n So n=0 n an(x − t) has radius of convergence R. 1 Second case: R = 1. The method in the case above works, after replacing R with zero. 1=n Third case: R = 0. Hence by Definition 21.2 the sequence fjanj g is not bounded above. Let K p=n p=n 1 be a positive number. Since limn!1 n = 1, there exists some N such that n ≥ 2 for all n ≥ N. p 1=n p 1=M Since fjn anj g is not bounded above, there exists some M > N such that jM aM j > 2K (for p 1=1 p 1=N p 1=n otherwise maxfj1 a1j ; ··· ; jN aN j ; 2Kg would be an upper bound for fjn anj g). Hence p 1=M p=M p 1=M 1 p 1=n jM aM j = M jM aM j > 2 2K = K. This shows that fjn anj g is not bounded above, P1 p n so n=0 n an(x − t) has radius of convergence 0. 28.1. Prove that the following series converge conditionally. Solution. (a) 1 + p1 − p2 + p1 + p1 − p2 + ··· 2 3 4 5 6 p1 Let fang be the sequence 1; 1; −2; 1; 1; −2;::: . Let bn = n . Note that the sequence of partial P1 sums of n=1 an is bounded and fbng is a decreasing sequence with limit 0. We apply Corollary 28.3 (Dirichlet's Test) to see that 1 1 2 1 1 2 X 1 + p − p + p + p − p + ··· = anbn 2 3 4 5 6 n=1 converges. P1 n (1−n)=n (b) n=1(−1) n n −1 1=n P1 Let an = (−1) n and let bn = n . Note that n=1 an converges by Theorem 25.1 (Alternat- ing Series Test) and that fbng is a bounded monotone sequence for n sufficiently large by Theorem 16.7. We apply Corollary 28.6 (Abel's Test) to see that 1 1 1 X n (1−n)=n X n −1 1=n X (−1) n = (−1) n n = anbn n=1 n=1 n=1 converges. 2 P1 28.3. Give an example of a convergent series n=1 an and a positive sequence fbng such that limn!1 bn = P1 0 and n=1 anbn diverges. ( (−1)n 0 n odd Solution. Let an = p and bn = . n p1 n n even P1 Then n=1 an converges by Theorem 25.1 (Alternating Series Test) and limn!1 bn = 0. Note P1 P1 1 n=1 anbn = m=1 2m diverges by Corollary 24.3. Pp 28.6. Let fang be a periodic sequence, an = an+p for all n and for fixed p, for which n=1 an = 0. Let P1 fbng be a decreasing sequence with limit 0. Prove that n=1 anbn converges. P1 Pp Solution. Consider the partial sums sn of the series n=1 an. Since fang is periodic and n=1 an = Pp 0, we see that sn+p = sn + i=1 an+i = sn for all n. Hence sn 2 fs1; :::; spg for all n. This shows P1 that the sequence of partial sums of the series n=1 an is bounded. Hence Corollary 28.3 (Dirichlet's P1 Test) applies, and we conclude n=1 anbn converges. P1 xn 29.3. Prove that n=0 n! converges absolutely for every x and 1 1 1 X xn X yn X (x + y)n ( )( ) = n! n! n! n=0 n=0 n=0 Solution. Note that xn+1 (n+1)! x lim n = lim = 0 n!1 x n!1 n! n + 1 so by the ratio test, the series converges absolutely for every x. xn yn By letting an = n! and bn = n! in Theorem 29.9, we get that 1 1 1 1 X xn X yn X X ( )( ) = ( a )( b ) n! n! n n n=0 n=0 n=0 n=0 1 X = cn n=0 1 n X X = ( aibn−i) n=0 i=0 1 n X X xi yn−i = ( ) i! (n − i)! n=0 i=0 1 n X 1 X n = ( xiyn−i) n! i n=0 i=0 1 X (x + y)n = n! n=0 where the last equality is by the binomial theorem. P1 P1 29.5. Give an example of convergent series n=0 an and n=0 bn whose Cauchy product diverges. Solution. 3 1 1 1 Let P a = P b = P (−1)n p 1 . This series converges by Theorem 25.1 (Alternating n=0 n n=0 n n=0 n+1 Series Test). The n-th term in the Cauchy product is n n n X X k n−k 1 1 n X 1 1 cn = akbn−k = (−1) (−1) p p = (−1) p p k + 1 n + k + 1 k + 1 n − k + 1 k=0 k=0 k=0 Since p 1 ≥ p 1 and likewise p 1 ≥ p 1 , we have k+1 n+1 n−k+1 n+1 n X 1 1 jc j ≥ = (n + 1) = 1 n n + 1 n + 1 k=0 P1 So limn!1 cn 6= 0, so the series n=0 cn diverges by Theorem 22.3. 4.