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math104-f21:hw6

HW 6

Due next Thursday, 10/7, 6pm

1. Ross Ex 14.1 (briefly describe your reasoning)

2. Ross Ex 14.4

3. Let n=1an\sum_{n=1}^\infty a_n be a series. Show that if m=1a2m\sum_{m=1}^\infty a_{2m} and m=1a2m1\sum_{m=1}^\infty a_{2m-1} both converges, then n=1an\sum_{n=1}^\infty a_n converges.

4. Show that if a series nan\sum_n a_n converges absolutely, then nanan+1\sum_n a_n a_{n+1} converges absolutely.

5. Give an example of divergent series nan\sum_n a_n of positive numbers ana_n, such that limnan+1/an=limnan1/n=1\lim_n a_{n+1} / a_n = \lim_n a_n^{1/n} = 1. And give an example of convergent series nbn\sum_n b_n of positive numbers bnb_n, such that limnbn+1/bn=limnbn1/n=1\lim_n b_{n+1} / b_n = \lim_n b_n^{1/n} = 1.

Solution

1. Ross 14.1. Determine which of the series converges and explain.

  • n4/2n\sum n^4/2^n, converges, by ratio test. an/an1=(n/(n1))4/21/2a_n / a_{n-1} = (n/(n-1))^4 / 2 \to 1/2
  • 2n/n!\sum 2^n/n!, converges, by ratio test. an/an1=2/n0a_n/a_{n-1} = 2/n \to 0.
  • n2/3n\sum n^2/3^n, converges, ratio test.
  • n!/(n4+3)\sum n!/(n^4+3), diverges, ratio test.
  • cos2n/n2\sum \cos^2 n/n^2, converges absolutely. By comparing with 1/n2\sum 1/n^2.
  • 1/logn\sum 1/\log n, diverges. By comparing with n1/n\sum_n 1/n.

2. Ross 14.4

  • 1/(n+(1)n)2\sum 1/(n + (-1)^n)^2 converges, by comparing with n=21/(n1)2\sum_{n=2}^\infty 1/(n-1)^2.
  • (n+1n)=1n+1+n12n\sum (\sqrt{n+1}-\sqrt{n}) = \sum \frac{1}{\sqrt{n+1}+\sqrt{n}} \geq \sum \frac{1}{2\sqrt{n}}, hence divergent. Or the partial sum sequence sn=n+1s_n = \sqrt{n+1}, and is divergent.
  • n!/nn\sum n!/n^n, we can do ratio test an+1/an=n+1(n+1)n+1/nn=1(1+1/n)n1/e<1a_{n+1} / a_n = \frac{n+1}{(n+1)^{n+1}/n^n} = \frac{1}{(1+1/n)^n} \to 1/e < 1 Hence it is convergent.

3. Problem: Let n=1an\sum_{n=1}^\infty a_n be a series. Show that if m=1a2m\sum_{m=1}^\infty a_{2m} and m=1a2m1\sum_{m=1}^\infty a_{2m-1} both converges, then n=1an\sum_{n=1}^\infty a_n converges.

Solution: Let bn=a2nb_n = a_{2n} and cn=a2n1c_n = a_{2n-1}, for n=1,2,n=1,2,\cdots. Let BN=n=1NbnB_N = \sum_{n=1}^N b_n and CN=n=1NcnC_N = \sum_{n=1}^N c_n, and AN=n=1NanA_N = \sum_{n=1}^N a_n. Then A2N=BN+CNA_{2N} = B_N + C_N, and A2N+1=BN+CN+1A_{2N+1} = B_N+C_{N+1}. Let B=limBN,C=limCNB = \lim B_N, C= \lim C_N. Hence limA2N\lim A_{2N} and limA2N+1\lim A_{2N+1} both exists and equals B+CB+C, thus limAN\lim A_N exists and equals B+CB+C.

4. Since nan\sum_n a_n converges, liman=0\lim a_n = 0, hence ana_n is bounded, say an<L|a_n|<L for some L>0L >0. Then nanan+1Lnan<\sum_{n} |a_n a_{n+1}| \leq L \sum_n |a_n| < \infty, hence converges absolutely.

5. divergent example n1/n\sum_n 1/n; convergent example, n1/n2\sum_n 1/n^2.

math104-f21/hw6.txt · Last modified: 2022/01/11 08:36 by pzhou