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--- math104-f21:hw6 [2021/10/01 21:38]
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 Due next Thursday, 10/7, 6pm
-. Calculate (a) $\lim (n!)^{1/n}$, (b) $\lim (n!)^{1/n}/n$
+. Ross Ex 14.1 (briefly describe your reasoning)
-. Show that if a series $\sum_n a_n$ absolutely converges, then $\sum_n a_n a_{n+1}$ converges absolutely.
+. Ross Ex 14.4
-. Ross Ex 14.1 (briefly describe your reasoning)
+. 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.
-. Ross 14.4
+. Show that if a series $\sum_n a_n$ converges absolutely, then $\sum_n a_n a_{n+1}$ converges absolutely.
-. 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 $\lim_n b_{n+1} / b_n = \lim_n b_n^{1/n} = 1$.
+. 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 $b_n$, such that $\lim_n b_{n+1} / b_n = \lim_n b_n^{1/n} = 1$.
+====== Solution ======
+. 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 $\sum_n 1/n$.
+. 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.
+. Problem: 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.
+Solution: Let $b_n = a_{2n}$ and $c_n = a_{2n-1}$, for $n=1,2,\cdots$. Let $B_N = \sum_{n=1}^N b_n$ and $C_N = \sum_{n=1}^N c_n$, and $A_N = \sum_{n=1}^N a_n$. Then $A_{2N} = B_N + C_N$, and $A_{2N+1} = B_N+C_{N+1}$. Let $B = \lim B_N, C= \lim C_N$. Hence $\lim A_{2N}$ and $\lim A_{2N+1}$ both exists and equals $B+C$, thus $\lim A_N$ exists and equals $B+C$.
+. Since $\sum_n a_n$ converges, $\lim a_n = 0$, hence $a_n$ is bounded, say $|a_n|<L$ for some $L >0$. Then $\sum_{n} |a_n a_{n+1}| \leq L \sum_n |a_n| < \infty$, hence converges absolutely.
+.  divergent example $\sum_n 1/n$; convergent example, $\sum_n 1/n^2$.

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