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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

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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.

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\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 $\sum_n\frac1{n^2}$.

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    $\begingroup$ This is an elegant answer +1 $\endgroup$ – Aryadeva Apr 17 at 15:27
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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.

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