Does this series approach an actual value? Does this series converge to an actual number for any value of $x$? $$\sum_ {k = 1}^{\infty} \frac{\ln(k)\sin( 2\pi kx) }{ k}. $$ I tried summing the series for $x=2/3$ on wolfram alpha, and it seems bounded, but it also just seems to oscillate alot. Does it actually approach something? If so, does anyone know how to speed up the convergence of a sequence like this?
 A: It does converge for all real $x$ by Dirichlet's test.  However, for almost all $k$ it only converges conditionally.  To speed up convergence a bit, you might try summation by parts.
EDIT: Let $$\eqalign{a_n &= \frac{\ln(n)}{n}\cr
d_n &= a_n - a_{n+1} = \frac{\ln(n)}{n} - \frac{\ln(n+1)}{n+1}\cr
b_n(x) &= \sin(2 \pi n x)\cr
B_n(x) &= \sum_{k=1}^n b_k(x) = \frac{\sin(2 \pi n x) - \sin(2 \pi (n+1)x) + \sin(2 \pi x)}{1 - \cos(2 \pi x)}}$$
Summation by parts says
$$ \sum_{k=1}^n a_k b_k = \sum_{k=1}^{n-1} d_k B_k + a_n B_n $$
and thus if $a_n \to 0$ and $B_n$ stays bounded as $n \to \infty$,
$$ \sum_{k=1}^\infty a_k b_k = \sum_{k=1}^\infty d_k B_k $$
Here is an animation of the partial sums $\sum_{k=1}^N a_k b_k$ (red) and $\sum_{k=1}^{N} d_k B_k$ (blue) 
: you can see that the blue curve settles down to a limit faster than the red curve does.

A: Note that
$$\sum_{k=1}^N \sin(2 \pi k x) = \csc(\pi x) \sin(N \pi x) \sin((N+1)\pi x)$$
Hence,
$$\left \vert \sum_{k=1}^N \sin(2 \pi k x) \right \vert = \left \vert \csc(\pi x) \sin(N \pi x) \sin((N+1)\pi x) \right \vert \leq \left \vert \csc(\pi x) \right \vert$$
Hence, $\displaystyle \left \vert \sum_{k=1}^N \sin(2 \pi k x) \right \vert$ is bounded for all $N$.
Also, we have that if $a_k = \dfrac{\log(k)}{k}$, then $a_k$ is a monotone decreasing sequence to $0$. Now use Dirichlet's test to conclude that the series converges for all $x$.
