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.
 A: \begin{align*}
\frac{(-1)^n}{n+\sin n}&=\frac{(-1)^n}{n}+\Bigl(\frac{(-1)^n}{n+\sin n}-\frac{(-1)^n}{n}\Bigr)\\
&=\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.

A: We have
$$\frac{(-1)^n}{n+\sin n}=\frac{(-1)^n}{n}\left(1+\frac{\sin n}{n}\right)^{-1}=\frac{(-1)^n}{n}\left(1-\frac{\sin n}{n}+o\left(\frac1n\right)\right)\\=\frac{(-1)^n}{n}+\frac{(-1)^{n+1}\sin n}{n^2}+o\left(\frac1{n^2}\right)$$
so the given series is convergent since it's sum of $3$ convergent series:


*

*the first by the Leibniz theorem

*the second the third by comparison with a Riemann series.

A: add two consecutive terms together,
$$\frac{1}{n+\sin n}-\frac{1}{n+1+\sin (n+1)}=\frac{\sin (n+1)-\sin n +1}{(n+\sin n)(n+1+\sin (n+1))}$$
Now do a compairison, 
$$\frac{|\sin (n+1)-\sin n +1|}{(n+\sin n)(n+1+\sin (n+1))}\leq \frac{3}{(n-1)n}$$
A: 
I am not very sure that my answer is correct or not as these tests are what i learned. Hope this may help you.
A: 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.
