Does the series $\sum\limits_{n=2}^\infty(-1)^n\ln\left(1+\frac{\sin n}{\ln n}\right)$ converge? I want to know the nature of the following series : $$\sum_{n=2}^{\infty}(-1)^n\ln\left(1+\frac{\sin(n)}{\ln(n)}\right)$$
So, I make an asymptotic expansion to $$(-1)^n\ln\left(1+\frac{\sin(n)}{\ln(n)}\right)=\frac{(-1)^n\sin(n)}{\ln(n)}+\frac{(-1)^{n+1}\sin^2(n)}{\ln^2(n)}+O\left(\frac{1}{\ln^2(n)}\right)$$ but $\sum_{n=2}^{\infty}O\left(\frac{1}{\ln^2(n)}\right)$ doesn't converge absolutely, and I can't group it with $\frac{(-1)^{n+1}\sin^2(n)}{\ln^2(n)}$ to search an equivalent because $\frac{(-1)^{n+1}\sin^2(n)}{\ln^2(n)}$ doesn't conserve the same sign. 
what should I do to know its nature?
 A: 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. 
