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HW 6
Due next Thursday, 10/7, 6pm
1. Ross Ex 14.1 (briefly describe your reasoning)
2. Ross Ex 14.4
3. Let $\sum_{n=1}^\infty a_n$ be a series. Show that if $\sum_{m=1}^\infty a_{2m}$ and $\sum_{m=1}^\infty a_{2m-1}$ both converges, then $\sum_{n=1}^\infty a_n$ converges.
4. Show that if a series $\sum_n a_n$ converges absolutely, then $\sum_n a_n a_{n+1}$ converges absolutely.
5. Give an example of divergent series $\sum_n a_n$ of positive numbers $a_n$, such that $\lim_n a_{n+1} / a_n = \lim_n a_n^{1/n} = 1$. And give an example of convergent series $\sum_n b_n$ of positive numbers $a_n$, such that $\lim_n b_{n+1} / b_n = \lim_n b_n^{1/n} = 1$.
Solution
1. Ross 14.1. Determine which of the series converges and explain.
$\sum n^4/2^n$, converges, by ratio test. $a_n / a_{n-1} = (n/(n-1))^4 / 2 \to 1/2$
$\sum 2^n/n!$, converges, by ratio test. $a_n/a_{n-1} = 2/n \to 0$.
$\sum n^2/3^n$, converges, ratio test.
$\sum n!/(n^4+3)$, diverges, ratio test.
$\sum \cos^2 n/n^2$, converges absolutely. By comparing with $\sum 1/n^2$.
$\sum 1/\log n$, diverges. By comparing with $\sum_n 1/n$.
2. Ross 14.4
$\sum 1/(n + (-1)^n)^2$ converges, by comparing with $\sum_{n=2}^\infty 1/(n-1)^2$.
$\sum (\sqrt{n+1}-\sqrt{n}) = \sum \frac{1}{\sqrt{n+1}+\sqrt{n}} \geq \sum \frac{1}{2\sqrt{n}}$, hence divergent. Or the partial sum sequence $s_n = \sqrt{n+1}$, and is divergent.
$\sum n!/n^n$, we can do ratio test $$a_{n+1} / a_n = \frac{n+1}{(n+1)^{n+1}/n^n} = \frac{1}{(1+1/n)^n} \to 1/e < 1$$ Hence it is convergent.