3. Let ∑n=1∞an be a series. Show that if ∑m=1∞a2m and ∑m=1∞a2m−1 both converges, then ∑n=1∞an converges.
4. Show that if a series ∑nan converges absolutely, then ∑nanan+1 converges absolutely.
5. Give an example of divergent series ∑nan of positive numbers an, such that limnan+1/an=limnan1/n=1. And give an example of convergent series ∑nbn of positive numbers bn, such that limnbn+1/bn=limnbn1/n=1.
Solution
1. Ross 14.1. Determine which of the series converges and explain.
∑n4/2n, converges, by ratio test. an/an−1=(n/(n−1))4/2→1/2
∑2n/n!, converges, by ratio test. an/an−1=2/n→0.
∑n2/3n, converges, ratio test.
∑n!/(n4+3), diverges, ratio test.
∑cos2n/n2, converges absolutely. By comparing with ∑1/n2.
∑1/logn, diverges. By comparing with ∑n1/n.
2. Ross 14.4
∑1/(n+(−1)n)2 converges, by comparing with ∑n=2∞1/(n−1)2.
∑(n+1−n)=∑n+1+n1≥∑2n1, hence divergent. Or the partial sum sequence sn=n+1, and is divergent.
∑n!/nn, we can do ratio test an+1/an=(n+1)n+1/nnn+1=(1+1/n)n1→1/e<1 Hence it is convergent.
3. Problem: Let ∑n=1∞an be a series. Show that if ∑m=1∞a2m and ∑m=1∞a2m−1 both converges, then ∑n=1∞an converges.
Solution: Let bn=a2n and cn=a2n−1, for n=1,2,⋯. Let BN=∑n=1Nbn and CN=∑n=1Ncn, and AN=∑n=1Nan. Then A2N=BN+CN, and A2N+1=BN+CN+1. Let B=limBN,C=limCN. Hence limA2N and limA2N+1 both exists and equals B+C, thus limAN exists and equals B+C.
4. Since ∑nan converges, liman=0, hence an is bounded, say ∣an∣<L for some L>0. Then ∑n∣anan+1∣≤L∑n∣an∣<∞, hence converges absolutely.
5. divergent example ∑n1/n; convergent example, ∑n1/n2.
math104-f21/hw6.txt · Last modified: 2022/01/11 08:36 by pzhou