Does the series $\sum\limits_{n=2}^\infty(-1)^n\ln\left(1+\frac{\sin n}{\ln n}\right)$ converge?

Solution 1:

This is an elaboration on my comment. The series converges. We apply the following lemma about the interchanging limits.

Lemma

A double sequence $\{a_{nk}\}$ satisfies

(1) $f(K):=\lim\limits_{N\rightarrow\infty} \sum\limits_{n=0}^N \sum\limits_{k=0}^K a_{nk}$ exists for each $K$.

(2) $g(N):=\lim\limits_{K\rightarrow\infty} \sum\limits_{k=0}^K \sum\limits_{n=0}^N a_{nk}$ converges uniformly in $N$.

(3) $A:=\lim\limits_{K\rightarrow\infty}\lim\limits_{N\rightarrow\infty} \sum\limits_{k=0}^K\sum\limits_{n=0}^N a_{nk}$ exists.

Then $\lim\limits_{N\rightarrow\infty}\lim\limits_{K\rightarrow\infty}\sum\limits_{k=0}^K \sum\limits_{n=0}^N a_{nk} = \lim\limits_{K\rightarrow\infty}\lim_\limits{N\rightarrow\infty} \sum\limits_{k=0}^K\sum\limits_{n=0}^N a_{nk}$. Thereby justifying the interchange of limits $$ \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} a_{nk} = \sum_{k=0}^{\infty}\sum_{n=0}^{\infty} a_{nk}. $$

Proof) Let $\epsilon>0$. The condition (3) means $\lim\limits_{K\rightarrow\infty}f(K)=A$. We need to prove $\lim\limits_{N\rightarrow\infty}g(N)=A$. By (2) and (3), we can find $K_0=K_0(\epsilon)$ such that $$ K\geq K_0 \Longrightarrow |f(K)-A|\le \epsilon, $$ $$ K\geq K_0 \Longrightarrow \left| g(N)-\sum_{k=0}^K \sum_{n=0}^N a_{nk} \right|\le \epsilon \ \ \textrm{ holds for all }N.$$ By (1), for each $K$, there exists $N_0=N_0(\epsilon,K)$ such that $$ N\ge N_0 \Longrightarrow \left|f(K)-\sum_{n=0}^N \sum_{k=0}^K a_{nk} \right| \le \epsilon. $$ If $N\ge N_0=N_0(\epsilon,K_0)$, then the above yields $$ |g(N)-A|\leq \left| g(N)-\sum_{k=0}^{K_0} \sum_{n=0}^N a_{nk} \right|+\left|\sum_{n=0}^N \sum_{k=0}^{K_0} a_{nk} -f(K_0)\right|+|f(K_0)-A|\le 3\epsilon. $$ Therefore, we have the result.

Main Problem

We consider the sum starting at $n=3$. By Taylor expansion of logarithm, $$ \sum_{n=3}^{\infty} (-1)^n \log \left( 1+ \frac{\sin n}{\log n}\right)=\sum_{n=3}^{\infty} (-1)^n \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}k \left(\frac{\sin n}{\log n}\right)^k. $$ We will prove that the interchanged sum $$ \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}k \sum_{n=3}^{\infty} (-1)^n \left(\frac{\sin n}{\log n}\right)^k. $$ converges, and it satisfies the conditions in the lemma. In the inner sum, let $$ a_n=\log^{-k} n, \ \ \ \, b_n=(-1)^n \sin^k n, \ \ \ \ B_m=\sum_{n=3}^m b_n. $$ Suppose we have the bound $|B_m|\leq M_k$. Since $a_n$ is decreasing (and summation by parts),
$$ \left|\sum_{n=3}^m a_nb_n\right| \leq M_k a_3=\frac{M_k}{\log^k 3}. $$ If we replace $3$ by any $n_0\geq 3$, this bound of $B_m$ together with Dirichlet's test can also be used in proving (1) in the Lemma.

By the power reduction formula for sine or Euler's formula applied to $\sin^k n$, there are $\alpha_{jk}$ such that $|\alpha_{jk}|\le 1$ and $$ \sin^k n = \left( \frac{e^{in} - e^{-in}}{2i}\right)^k=\sum_{j=-k}^k \alpha_{jk} e^{ijn}=\alpha_{0k} + \sum_{0<|j|\leq k} \alpha_{jk} e^{ijn}.$$ Multiplying $(-1)^n$ and summing over $n$ gives $$ |B_m| = \left|\sum_{n=3}^m (-1)^n\left(\alpha_{0k}+ \sum_{0<|j|\le k} \alpha_{jk} e^{ijn}\right)\right|\le 1+ \sum_{0<|j|\le k}\left| \sum_{n=3}^m e^{i(j+\pi )n}\right|$$ $$\le 1+ \sum_{0<|j|\le k} \frac2{|1-e^{i(j+\pi)}|}. $$ The magnitude of the sum over $j$ on the right depends on how close we can approximate $j+\pi$ mod $2\pi$ to $0$ mod $2\pi$. By the result on the irrationality measure of $\pi$, we obtain that there is an absolute constant $C'>0$ with $$ \frac2{|1-e^{i(j+\pi)}|} \le C' j^7 \ \ \textrm{ for }0<|j|\le k. $$ Thus, there is an absolute constant $C>0$ with $$ |B_m|\leq M_k:= Ck^8. $$

Since the sum $\sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1}}k \frac{Ck^8}{\log^k 3}$ is absolutely convergent, we have (2) and (3) in the Lemma. Since (1) was verified, we may apply the lemma and the sum converges.