Does the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n(\sin(n)+2)}$ converge or diverge? I was thinking about it and was stumped. Mathematica claims it converges.
 A: My claim is as follows:

Claim. For any $\theta \in \Bbb{Q}$, we have
  $$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n(2+\sin n\theta)} = -\frac{\log 2}{\sqrt{3}} - \frac{2}{\sqrt{3}} \Re \sum_{n=1}^{\infty} \frac{i^n}{(2+\sqrt{3})^n} \log(1+e^{in\theta}). \tag{*} $$

For example, if we take $\theta = 1$ as in the original problem and evaluating the first 10000 terms, we get
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n(2+\sin n)} \approx -0.27206008766393670467\cdots, $$
which agrees well with MathUser's observation.

Before the proof (optional). 
My first trial was to use some equidistribution results, but this turned out to be daunting because I know nothing about asymptotic behavior of
$$ \frac{1}{n}\sum_{k=1}^{n} \frac{1}{2+\sin k} = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{2+\sin x} + \boxed{\epsilon_n : \text{error term}}, $$
since we can write
$$ \sum_{n=1}^{N} \frac{(-1)^n}{n(2+\sin n)}
= 2\sum_{n=1}^{N} (-1)^{n} \epsilon_n + (-1)^N \epsilon_N + \sum_{n=1}^{N} \frac{(-1)^n}{n} (C + \epsilon_n)$$
with the constant $C = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{2+\sin x}$. So I gave up this approach and tried new one, which indeed led me to a correct proof. I consider my proof a sledgehammer method, though, and I suspect that there is a simpler proof.

Preliminary. We first introduce two big theorems. The first one is the following famous version of Tauberian theorems:

Littlewood's Tauberian Theorem. If $a_n = \mathcal{O}(1/n)$ and $\sum a_n$ is Abel summable, i.e.,
  $$ \lim_{s \to 0^+} \sum_{n=1}^{\infty} a_n e^{-ns} = S$$
  converges, then $\sum a_n$ converges in the usual sense and
  $$ \sum_{n=1}^{\infty} a_n = S. $$

The next theorem deals with how $e^{in\theta}$ gets closer to $-1$ as $n$ grows.

Theorem. The irrationality measure of $1/\pi$ is finite. In particular, there exists $c, \mu > 0$ such that
  $$ \mathrm{dist}(n, \pi \Bbb{Z}) \geq c n^{-\mu} $$
  for any $n = 1, 2, \cdots $.


Proof. The actual proof is quite straightforward. In view of the Littlewood's Tauberian theorem, it suffices to show that (*) holds in Abel summbability sense. To ghis end, let $r = i(2-\sqrt{3})$ and notice that for $x \in \Bbb{R}$,
\begin{align*}
\frac{1}{2+\sin x}
&= \frac{1}{\sqrt{3}} \left( \frac{1}{1 - re^{ix}} + \frac{\bar{r}e^{-ix}}{1 - \bar{r}e^{-ix}} \right) \\
&= \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} \Re \sum_{k=1}^{\infty} r^k e^{ikx}.
\end{align*}
So by the Fubini's theorem, if $s > 0$, we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{(-1)^n}{n(2+\sin n\theta)}e^{-ns}
&= \sum_{n=1}^{\infty} \frac{(-1)^n}{n}e^{-ns} \left( \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} \Re \sum_{k=1}^{\infty} r^k e^{ikn\theta} \right) \\
&= -\frac{\log (1 + e^{-s})}{\sqrt{3}} - \frac{2}{\sqrt{3}} \Re \sum_{k=1}^{\infty} r^k \log(1 + e^{-s+ik\theta}).
\end{align*}
Now the Theorem 2 shows that
$$ \left| \log(1 + e^{-s}e^{ik\theta}) \right|
\leq -\log|\sin k\theta| + \mathcal{O}(1)
\leq C \log k + \mathcal{O}(1) \tag{1} $$
uniformly in $k$ and $s$. So we can take termwise limit as $s \to 0^+$ and the resulting series
$$ -\frac{\log 2}{\sqrt{3}} - \frac{2}{\sqrt{3}} \Re \sum_{k=1}^{\infty} r^k \log(1 + e^{ik\theta}) $$
still converges absolutely. This shows that (*) is true in Abel summability sense and hence completes the proof of our claim. ////

Addendum: Proof of (1). I admit that I omitted some detail when proving (1). So here is my derivation: Notice first that
\begin{align*}
\left| \log(1 + e^{-s}e^{ik\theta}) \right|
&= \left| \log\left| \frac{1 + e^{-s}e^{ik\theta}}{2} \right| + i\arg(1 + e^{-s}e^{ik\theta}) + \log 2 \right| \\
&\leq - \log\left| \frac{1 + e^{-s}e^{ik\theta}}{2} \right| + \pi + \log 2 \\
&= - \log\left| 1 + e^{-s}e^{ik\theta} \right| + \pi + 2\log 2
\end{align*}
where the intermediate inequality follows from the triangle inequality together with the observation that $|1 + e^{-s}e^{ik\theta}| \leq 2$. Now let us write $r = p/q$ for $p, q \in \Bbb{Z}$ and $q > 0$. Then
\begin{align*}
\left| 1 + e^{-s}e^{ik\theta} \right|
&\geq \left| \Im (1 + e^{-s}e^{ik\theta}) \right|
 = e^{-s}\left| \sin(k\theta) \right|
 \geq e^{-s} \cdot \frac{2}{\pi}\,\mathrm{dist}(k\theta, \pi \Bbb{Z}) \\
&= \frac{2}{\pi q}e^{-s}\,\mathrm{dist}(kp, q \pi \Bbb{Z})
 \geq  \frac{2}{\pi q}e^{-s}\,\mathrm{dist}(k|p|, \pi \Bbb{Z}) \\
&\geq  \frac{2c}{\pi|p|^{\mu}q}e^{-s} k^{-\mu}.
\end{align*}
Taking negative log, we get
$$ -\log \left| 1 + e^{-s}e^{ik\theta} \right| \leq \mu \log k + s + \mathcal{O}(1). $$
As we are interested only in the limit as $s \to 0^+$, by restricting the range of $s$ onto $0 < s \leq 1$, the term $s$ can be absorbed into the $\mathcal{O}(1)$ term. Therefore this yields the desired inequality (1) with the constant $C = \mu$.
A: A non-answer that might shed some light. Added here because comments do not allow for images. This picture shows the value of the sum $f(m) = \sum_{n=1}^{m} \frac{(-1)^n}{n(\sin(n)+2)}$ with $m$ starting in $0$ and ending in $100,1000,10000$. 
Notice the change in the y-scale axis. I am almost sure it converges, can anyone guess (or find) the value of the lower and upper bounds? Maybe using a comparison test and sandwich law you can determine the number to which it actually converges.
 
