For what real numbers $a$ does the series $\sum \frac{\sin(ka)}{\log(k)}$ converge or diverge? For what values ​​of $a\in \mathbb{R}$ does this series converge or diverge? 
$$\sum \frac{\sin(ka)}{\log(k)}$$
Clearly for $a=\pi n$ with $n\in \mathbb{Z}$ the series converges because $\sin(ka)=0$, but I do not know how to continue. 
 A: You can conclude it based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.
Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.
First note that from Abel summation, we have that
$$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N b(n)(A(n)-A(n-1)) = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\
= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1} b(n+1)A(n) = b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that
$$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$
Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n)  (b(n)-b(n+1))$$ converges absolutely. Now since
$$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges.
In your case, $a(n) = \sin(an)$. Hence, $$A(N) = \displaystyle \sum_{n=1}^N a(n) = \dfrac{\sin(a(N+1)/2) \sin(aN/2)}{\sin(a/2)} \leq \csc(a/2)$$ when $a\neq 2m \pi$ is bounded. If $a=2 m \pi$, then $a(n)$ is trivially $0$ and hence again it is bounded.
 Also, $b(n) = \dfrac1{\log(n)}$ is a monotone decreasing sequence converging to $0$.
Hence, we have that $$\sum_{n=1}^N \dfrac{\sin(an)}{\log n}$$ converges for all $a \in \mathbb{R}$.
