convergence of the series $ \sum_{n=1}^{\infty}\frac{\sin n}{n+2\cos n} $ I know that the series $ \sum_{n=1}^{\infty}\frac{\sin n}{n+2\cos n} $ is not absolutely convergent, but I don't know how to determine if its conditional convergent or divergent. I can't use any comparison tests because its not a positive series, I can't use Dirichlet's or Abel's tests since $ n+2\cos n $  is not monotonic series. Any ideas will help. Thanks 
 A: \begin{eqnarray}
\frac{\sin n}{n+2\cos n}
&=&
\frac{\sin n}n+\left(\frac{\sin n}{n+2\cos n}-\frac{\sin n}n\right)
\\
&=&
\frac{\sin n}n-\frac{2\cos n\sin n}{n(n+2\cos n)}\;.
\end{eqnarray}
The sum over the first term is known to converge (which you can show e.g. using the Dirichlet test), and the second term can be bounded by the terms of the convergent series $\sum_n\frac1{n^2}$.
A: You have that
$$
a_n  = \frac{{\sin n}}
{{n + 2\cos n}} = \frac{{\sin n\left( {n - 2\cos n} \right)}}
{{n^2  - 4\cos ^2 n}} = \frac{n}
{{n^2  - 4\cos ^2 n}}\sin n - \frac{{\sin 2n}}
{{n^2  - 4\cos ^2 n}}
$$
Now it easy to prove that 
$$
\frac{n}
{{n^2  - 4\cos ^2 n}}
$$
is monotonically decreasing to $0$ thus
$$
\sum\limits_{n = 1}^{ + \infty } {\frac{n}
{{n^2  - 4\cos ^2 n}}\sin n} 
$$ 
is convergent by Dirichlet's test. On the other side
$$
\sum\limits_{n = 1}^{ + \infty } {\frac{{\sin 2n}}
{{n^2  - 4\cos ^2 n}}} 
$$
is absolutely convergent. Thus your series is convergent.
A: First of all :
\begin{aligned} \left(\forall n\in\mathbb{N}^{*}\right),\ \frac{\sin{n}}{n+2\cos{n}}=\frac{\sin{n}}{n}\times\frac{1}{1+\frac{2\cos{n}}{n}}&=\frac{\sin{n}}{n}-\frac{1}{n^{2}}\times\frac{\sin{\left(2n\right)}}{1+\frac{2\cos{n}}{n}}\\ &=\frac{\sin{n}}{n}+v_{n} \end{aligned}
With $ v_{n}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) \ \ \ \left(*\right) $
Let $ n\in\mathbb{N}^{*} $, denoting $ A_{n}=\sum\limits_{k=1}^{n}{\sin{k}} $, we have : \begin{aligned} \sum_{k=1}^{n}{\frac{\sin{k}}{k}}&=\sin{\left(1\right)}+\sum_{k=2}^{n}{\frac{A_{k}-A_{k-1}}{k}}\\ &=\sin{\left(1\right)}+\sum_{k=2}^{n}{\frac{A_{k}}{k}}-\sum_{k=2}^{n}{\frac{A_{k-1}}{k}}\\ &=\sum_{k=1}^{n}{\frac{A_{k}}{k}}-\sum_{k=1}^{n-1}{\frac{A_{k}}{k+1}}\\ \sum_{k=1}^{n}{\frac{\sin{k}}{k}}&=\sum_{k=1}^{n-1}{\frac{A_{k}}{k\left(k+1\right)}}+\frac{A_{n}}{n} \end{aligned}
Since $ \left\lbrace A_{n}\right\rbrace_{n} $ is bounded, $ \sum\limits_{n\geq 1}{\frac{A_{n}}{n\left(n+1\right)}} $ converges, and thus $ \sum\limits_{n\geq 1}{\frac{\sin{n}}{n}} \cdot $
Using $ \left(*\right) $, $ \sum\limits_{n\geq 1}{v_{n}} $ also converges.
Hence, $ \sum\limits_{n\geq 1}{\frac{\sin{n}}{n+2\cos{n}}} $ converges.
