A curious convergence condition 
How do you prove there exists an $M \in \mathbb{R}^+$ such that: 
  $$\left|\frac{-1}{\sin(1)+2} + \frac{1}{\sin(2)+2} +...+ \frac{(-1)^N}{\sin(N)+2}\right| \le M$$ for all $N \ge 1$?

This is the conditon for Dirichlet's test to determine convergence of: $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n(\sin(n)+2)},$$ which Mathematica claims to converge.
The given bound seems likely to hold based on a WolframAlpha search: http://m.wolframalpha.com/input/?i=sum+from+n+%3D+1+to+m+of+%28-1%29%5En%2F%28sin%28n%29%2B2%29&x=0&y=0
If you click "more terms" a few times, the graph will look periodic, with period about $44$, which is close to $14\pi$. Coincidence? Maybe.
 A: This problem can be approached in the following way. Over $[0,2\pi]$, we have:
$$\frac{1}{2+\sin x}= \frac{1}{\sqrt{3}}+\sum_{k\geq 0}a_k \sin((2k+1)x)+\sum_{k\geq 1}b_k \cos(2kx)\tag{1}$$
where:
$$ a_{k}=\frac{1}{\pi}\int_{0}^{2\pi}\frac{\sin((2k+1)x)}{2+\sin x}\,dx,\qquad b_k=\frac{1}{\pi}\int_{0}^{2\pi}\frac{\cos(2kx)}{1+2\sin x}\,dx.\tag{2}$$
Since:
$$\sum_{k=1}^{N}(-1)^k e^{ikx} = \frac{e^{ix}}{1+e^{ix}}\left(e^{iN(\pi+x)}-1\right) $$
it follows that:
$$ \max\left(\left|\sum_{k=1}^{N}(-1)^k \cos(kx)\right|,\left|\sum_{k=1}^{N}(-1)^k \sin(kx)\right|\right)\leq\frac{1}{\left|\cos\frac{x}{2}\right|}\tag{3}$$
hence if $f(x)=\frac{1}{2+\sin x}$ we have:
$$ \left|\sum_{k=1}^{N}(-1)^k f(k)\right| \leq \frac{1}{\sqrt{3}}+\sum_{n\geq 1}\frac{\left|a_n\right|+\left|b_n\right|}{\left|\cos\frac{n}{2}\right|}\tag{4}$$
so the LHS is bounded because the irrationality measure of $\pi$ is finite while $f(x)$ is a $C^\infty$ function over $[0,2\pi]$, giving that the RHS of $(4)$ is a convergent series.
