Determining convergence or divergence of series $ \sum_{n=1}^{\infty} \frac{(-1)^n}{ n+\sin (n)} $ I am wondering the convergence or divergence following series

$$
\sum_{n=1}^{\infty} 
\frac{(-1)^n}{ n+\sin (n)}
\\
$$

My 1st attempt is 'alternating series test' 

$$
$$

But, $$\frac{1}{n+\sin(n)}$$

isn't monotone decreasing. SO I failed...

$$
$$

Please give me some advice.

Thanks in advance.
\begin{align*}
\frac{(-1)^n}{n+\sin n}&amp;=\frac{(-1)^n}{n}+\Bigl(\frac{(-1)^n}{n+\sin n}-\frac{(-1)^n}{n}\Bigr)\\
&amp;=\frac{(-1)^n}{n}-\frac{(-1)^n\sin n}{(n+\sin n)n}.
\end{align*}


$\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{n}$ converges conditionally.
$\displaystyle\sum_{n=1}^\infty\frac{(-1)^n\sin n}{(n+\sin n)n}$ converges absolutely.
Easier: see that $-1 \leq \sin n \leq 1 $, hence the bounds on the summand are $\frac{(-1)^n}{n-1}$ and $\frac{(-1)^n}{n+1}$ which converge by alternating test (Leibniz test) and hence your original series converges.