For what values of $a$, $\sum_{n=1}^{\infty}\frac{(\cos n)(\sin na)}{n}$ converges? For what values of $a$, $$\sum_{n=1}^{\infty}\dfrac{(\cos n)(\sin na)}{n}$$ converges? Every hint is appreciated.
I know that $(\cos n)(\sin na)=\dfrac{1}{2}(\sin (n+1)a+\sin (n-1)a)$. 
 A: Kobe's answer perfectly deals with convergence. As a side note, it is interesting to point out that
$$ f(x)=\sum_{n\geq 1}\frac{\sin(nx)}{n} $$
is the Fourier series of the function $\frac{\pi-x}{2}$ over the interval $(0,2\pi)$, hence
$$ g(x)=\sum_{n\geq 1}\frac{\cos n\sin(nx)}{n} $$
is a $2\pi$-periodic function whose value is given by:
$$ g(x)=\left\{\begin{array}{rcl}-\frac{x}{2}&\text{if}&x\in[0,1),\\\frac{\pi-2}{4}&\text{if}&x=1,\\\frac{\pi-x}{2}&\text{if}&x\in(1,2\pi-1),\\\frac{2-\pi}{4}&\text{if}&x=2\pi-1,\\\pi-\frac{x}{2}&\text{if}&x\in(2\pi-1,2\pi],\end{array}\right.$$
so $g(x)$ is everywhere defined and $|g(x)|<\frac{\pi-1}{2}.$
A: Assuming $a$ is real, the series converges for all $a$. It follows from the fact that $\sum_{n = 1}^\infty \sin[n(a+1)]/n$ and $\sum_{n = 1}^\infty \sin[n(a-1)]/n$ converge for all $a$. Let's consider the latter series. If $a - 1$ is an integral multiple of $2\pi$, then each summand is $0$, so the series converges to $0$. If $a-1$ is not an integral multiple of $2\pi$, then the series converges by Dirichlet's test. To see this, note that for $N \ge 1$, \begin{align}\sum_{n = 1}^{N} \sin[n(a-1)] &= \csc\left(\frac{a-1}{2}\right) \sum_{n = 1}^{N} \sin[n(a-1)]\sin\left(\frac{a-1}{2}\right)\\
&= \csc\left(\frac{a-1}{2}\right) \sum_{n = 1}^{N-1} \left(\cos\left[\left(n-\frac{1}{2}\right)(a-1)\right] - \cos\left[\left(n + \frac{1}{2}\right)(a-1)\right]\right)\\
&= \csc\left(\frac{a-1}{2}\right) \left\{\cos\left(\frac{a-1}{2}\right) - \cos\left[\left(N + \frac{1}{2}\right)(a-1)\right]\right\}.
\end{align}
(Keep in mind that $\csc((a - 1)/2)$ is defined since $(a - 1)/2$ is not an integral multiple of $\pi$.) It follows that the sequence of partial sums of the series $\sum_{n = 1}^\infty \sin[n(a-1)]$ is bounded by $2|\csc((a-1)/2)|$. Furthermore, the sequence $\{\frac{1}{n}\}_{n = 1}^\infty$ decreases to $0$. Thus, by the Dirchlet test, the series $\sum_{n = 1}^\infty \sin[n(a-1)]/n$ converges. A similar argument shows that the series $\sum_{n = 1}^\infty \sin[n(a+1)]/n$ converges for all $a$.
