Study the convergence of the series: $\sum_{n=2}^{\infty} \frac{\cos(nx)\sin(\frac{x}{n})}{\ln n}$ I am studying the convergence of the following series:
$$\sum_{n=2}^{\infty}\frac{\cos(nx)\sin\frac{x}{n}}{\ln n}$$
I thought about using the Dirichlet's Test, according to which:
If we have a series of the form $\sum_{n=0}^{\infty}a_nb_n$ with
$\,\,(i)\,\, a_n$ is decreasing and tends to $0$, and
$(ii)\,\, t_n = b_0 + b_1 +\cdots+ b_n$ is bounded,
then $\sum_{n=0}^{\infty}a_nb_n$ is convergent.
Now, my only problem is that I do not really know how to choose $a_n$ and $b_n$. 
I though about choose $a_n = ln\space n$ and $b_n=\cos(nx)\space \sin\frac{x}{n}$ but I do not really know if I can calculate the sum for $b_n$.
Can you help me find out how to solve this?
 A: A. If $x\ne 2k\pi$, then, 
$$
c_n=\sin\Big(\frac{x}{n}\Big)
$$
eventually maintains sign, say $\sigma$ is its eventual sign, and its absolute value is strictly decreasing and tends to zero. In particular
$$
b_n=\frac{\sigma\sin\big(\frac{x}{n}\big)}{\ln n},
$$ 
is eventually strictly decreasing and tends to zero.
Meanwhile, 
$$
A_n=\cos x+\cos 2x+\cdots+\cos nx=\mathrm{Re}\big(\mathrm{e}^{xi}+
\mathrm{e}^{2xi}+\cdots+\mathrm{e}^{nxi}\big)
=\mathrm{Re}\left(\frac{\mathrm{e}^{(n+1)xi}-\mathrm{e}^{xi}}{\mathrm{e}^{xi}-1}\right)
$$ 
and thus
$$
|A_n|\le \frac{2}{|\mathrm{e}^{xi}-1|}=\frac{2}{|\sin(x/2)|},
$$
and
hence $A_n$ is bounded, whenever $x\ne2k\pi$. Then convergence is a consequence of Dirichlet's Test.
B. If $x=2k\pi$ and $x\ne 0$, then our series looks like
$$
\sum_{n=2}\frac{\sin (\frac{x}{n})}{\ln n}
$$
in which case we obtain divergence via Limit Comparison Test, since
$$
\frac{\displaystyle\frac{\sin (\frac{x}{n})}{\ln n}}{\displaystyle\frac{1}{n\ln n}}=n\sin\big(\frac{x}{n}\big)\to x.
$$
C. If $x=0$, then the sum is equal to zero.
