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Solutions to Assignment 6, Math 220 1 For each of the following series, indicate whether they converge or diverge and give reasons for your answers. P 1 a: ∞ k 2 +k Pk=1 −k b: P ∞ k=1 k ∞ 1 c: k=2 k log k P∞ d: P k=1 sin(π/k) 1 e: ∞ k=3 k 2 −3k+2 P∞ f: k=1 kk!k Solution P 1 a: Converge by comparison to ∞ k=1 k 2 b: Converge by the root test. c: Diverge by the integral test. P 1 d: Diverge by comparison with ∞ P∞k=1 2k e: Converge by comparison to k=3 k2 f: Converge by, say, the ratio test 2 For each of the following series, indicate whether they converge or diverge and give reasons for your answers. Which converge absolutely? P (−1)k √ a: ∞ k=1 k P∞ k b: k=1(−1)k k+1 P∞ c: k=1 (−1)k k2k+1 3 −4 P∞ k 2k k 2 d: k=1(−1) k! Solution a: Converge by the alternating series test, conditionally by the integral test. k b: Diverge since (−1)k k+1 fails to approach 0 P 3 c: Converge absolutely by comparison to ∞ k=3 k 2 d: Converge absolutely by the ratio test P P 2 3 If an with an > 0 is convergent, then is an always convergent? Either prove it or give a counterexample. Pn Pn 2 Solution This is true. If we let sn = k=1 ak and Sn = k=1 ak , then the triangle inequality tells us that Sn ≤ s2n . Since lim sn exists it follows that sn is bounded above and hence, so is SnP . Since Sn is monotone and bounded, it is convergent and thus, so is a2n . 1 2 P P√ 4 If an with an > 0 is convergent, then is an always convergent? Either prove it or give a counterexample. Solution A counterexample follows by taking an = 1 . n2 P 5 Can youPgive an example of a convergent series xn and a diverP gent series yn such that (xn + yn ) is convergent. Solution The answer isPno. If so does their difference yn . P P (xn + yn ) and xn converge, then P 6 Find an explicit expression for the nth partial sum of ∞ n=2 log(1 − 2 1/n ) to show that this series converges to − log 2. Is this convergence absolute? P n+1 2 . We prove Solution We claim that N log(1 − 1/n ) = log n=2 2n this by induction. Since N = 2 implies that the sum is just log(3/4), we may suppose that the claim is true for N = k ≥ 2. Then 2 k+1 X k+1 k + 2k 2 log(1 − 1/n ) = log + log 2k k 2 + 2k + 1 n=2 and the result follows by simple algebra. P∞ 2 The above formula tells us that n=2 log(1 − 1/n ) converges to log(1/2) = − log 2. Since all the terms in the series are negative, P∞ 2 n=2 | log(1 − 1/n )| thus converges also, to log 2. 7 Show that the series 1 1 1 1 1 1 1 1 1+ − + + − + + − +··· 2 3 4 5 6 7 8 9 is divergent. Solution If we let sn denote the partial sums of the above series, we have that 1 1 1 1 1 1 1 1 s3n = 1 + − + − + − + + · · ·+ . 2 3 4 5 6 3n − 2 3n − 1 3n It follows that 1 1 1 s3n > 1 + + + · · · + 4 7 3n − 2 and so 1 1 1 1 s3n > 1+ + +···+ . 3 2 3 n P 1 Since ∞ n=1 n diverges, it follows that s3n (and thus sn ) does likewise.