Q:

$\sum\limits_{n=1}^{\infty}\frac{\sin(nx)}{{(n^4+x^4)}^{\frac{1}{3}}}$ How to major $\sum\limits_{n=1}^{\infty}{\sin(nx)}$?

Problem: Show that the following series $\sum\limits_{n=1}^{\infty}\frac{\sin(nx)}{{(n^4+x^4)}^{\frac{1}{3}}}$ is convergent.
It was expressly indicated that I should prove the series convergence in the following way:

Dirichlet´s test for Uniform convergence: The series converge uniformly
$\sum_\limits{n=k}^{\infty}f_ng_n$
on $S$
if $\{f_n\}$ converges uniformly to zero on $S$, $\sum(f_{n+1}-f_n)$ converges absolutely uniformly on $S$, and
$||g_k+g_{k+1}+...||_S\leqslant M\:\:n\geqslant K$
for some constant $M$.

Resolution attempt:
$f_n=\frac{1}{{(n^4+x^4)}^{\frac{1}{3}}}\to 0$ as $n\to\infty$
Now I must prove  $\sum\limits_{n=1}^{\infty}{\sin(nx)}\leqslant M$
$\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}$
Question:
How can I use $\frac{e^{inx}-e^{-inx}}{2i}$ to major $\sum\limits_{n=1}^{\infty}{\sin(nx)}$?
Thanks in advance!

A:

From $\cos (A\pm B)=\cos A \cos B \mp \sin A \sin B$ we have $\cos (A-B)-\cos (A+B)=2\sin A \sin B.$
Therefore (with $h(x,n)$ being an abbreviation) we have $$[1]\quad 2\sin mx \sin (x/2)=  h(x,m) =\cos (m-1/2)x-\cos (m+1/2)x. $$ 
Let $f(x,n)=\sum_{m=1}^n \sin mx.$ Then $$f(x,n)\sin (x/2)=\frac {1}{2}\sum_{m=1}^n  h(x,m).$$  From the RHS of $[1]$  this is a telescoping sum, so $$f(x,n)\sin (x/2)=\cos (x/2)-\cos (n+1/2)x.$$   If $\sin (x/2)\ne 0 \;$ (that is, if $\frac {x}{2\pi} \not \in \Bbb Z $ ) then $$f(x,n)=\cot (x/2)-\frac {\cos (n+1/2)x} {\sin  (x/2)}.$$
From this I see that my comment that $\{f(x,n): n\in \Bbb N\}$ may be unbounded for some real $x$ is false. (Sorry about that.).
However if $0\ne ix\in \Bbb R$ then $\cos (n+1/2)x=\cosh (n+1/2)x \to \infty$ as $n\to \infty.$
For the (trivial) case $\sin (x/2)=0$ we have $\sin nx=0$ for all $n\in \Bbb N$ and $f(x,n)=0.$
For $x\in \Bbb R$ we may obtain the  closed formula for $f(x,n)$  by observing that it is the imaginary part of $g(x,n)=\sum_{m=1}^n e^{mix}.....$ If $e^{ix/2}\ne 1$ then $$g(x,n) =e^{ix}\cdot \frac {e^{nix}-1}{e^{ix}-1}=$$ $$=\frac {e^{ix/2}}{2i\sin (x/2)}\cdot e^{nix/2}\cdot 2i\sin (nx/2)=$$ $$=e^{i(n+1/2)x}\cdot\frac {\sin (nx/2)}{\sin  (x/2)}.$$ 

