Math 312 Intro to Real Analysis Homework 5 Solutions Stephen G Simpson Friday March 20 2009 The assignment consists of Exercises 14 3 14 4 14 6 14 13 15 3 15 4 15 7 in the Ross textbook Each problem counts 10 points P In solving some of these exercises we use the fact that 1 ns is convergent if s 1 and divergent if s 1 This result appears as Theorem 15 1 in the textbook X 1 1 an 1 converges by the ratio test 14 3 a 0 an n 1 n X 2 cos n converges by comparision with the geometric series b 3n X 3 1 1 3 1 because 1 2 cos x 3 n 3 9 2 P 3 n c 1 2 n converges by comparison with the geometric series P 1 2n because 2n n 2n for all n P 1 n 50 n2 converges by comparison with the geometric series d 2 P n 52 21 because 50 n2 52 P e sin n 9 diverges by Corollary 14 5 because lim sin n 9 does not exist X 100n 100 an 1 f converges by the ratio test 0 n an n 1 P 14 4 a 1 n 1 n 2 converges by comparison with the convergent series P 1 n 1 2 because n 1 n 2 n 1 2 for all n P P n 1 n 1 n 1 n diverges by comparison b P with the divergent series 1 2 n 1 because n 1 n 2 n 1 for all n n X n n an 1 1 c converges by the ratio test 1 nn an n 1 e P 14 6 Assume that an converges and that bn is a bounded sequence Let M be such that bn M for all n Then an bn an bn M an 1 P Hence an bn Pis convergent Pby comparison with the convergent series P M an M an P Thus an bn is absolutely convergent It follows by Corollary 14 7 that an bn is convergent P P 2 n 14 13 a n 0 arn where a 32 and r 23 By equation 2 on n 1 3 2 n P page 91 the sum of this geometric series is n 1 32 3 2 2 1 3 2 P n 3 Similarly with a r 32 we have n 1 32 25 1 23 b By cancellation the nth partial sum is Pn 1 Pn 1 1 k 1 k k 1 k 1 k k 1 1 1 1 12 21 31 n 1 n1 n1 n 1 1 1 n 1 for all n Taking the limit as n we obtain c By cancellation the nth partial sum is Pn k 1 Pn k 1 k k 1 2k 1 k 1 2k 2k 1 2 3 2 1 2 4 4 8 1 2 1 k 1 k k 1 P n 1 2n 1 n 1 2n 1 for all n Taking the limit as n we obtain d Using the result of part c we have P n 21 42 83 n 1 2n 4 1 8 4 P 4 2 16 n 1 n 1 2n 1 1 2 3 32 n 2n k 1 k 1 2k 1 P 1 n 2n n 1 2n 1 12 2 15 3 Letting u log x we have du dx x For s 6 1 we have Z Z u s 1 du dx s x log x s s 1 log 2 log 2 u 2 and this is finite if s 1 infinite if s 1 Thus by the integral test we X 1 converges if s 1 and diverges if s 1 For the see that n log n s n 2 special case s 1 we have Z Z dx du log u log 2 x log x u 2 log 2 2 and this is infinite hence X 1 diverges by the integral test n log n n 2 P 1 n log n diverges by comparison with the divergent series 1 n This is because log n n hence n log n n P P b log n n diverges by comparison with the divergent series 1 n 15 4 a P This is because log n n 1 n for all n c Letting u log log x we have du dx x log x hence Z Z dx du log u log log 4 x log x log log x 4 log log 4 u P hence 1 n log n log log n diverges by the integral test P P d log n n2 converges by comparison with the convergent series 1 n3 2 This is because log n n hence log n n2 1 n n 1 n3 2 P 15 7 Assume that an is a nonincreasing sequence of real numbers and an converges We are going to prove that lim nan 0 Note first that an 0 for all n This is because if an 0 for some n then since our sequence is nonincreasing we would have am an 0 for all m n hence lim am 6 0 which would contradict Corollary 14 5 Now given 0 apply the Cauchy criterion to find an N so large that am am 1 an whenever N m n In particular we have aN 1 aN 2 an and aN 1 aN 2 an and n n N 2 whenever n 2N From these observations it follows that nan n N an 2 whenever n 2N This shows that lim nan 2 0 and it is then immediate that lim nan 0 Q E D 3