Show that $ x\cdot\cos(x)+\sin(x)/2=\sum_{n=2}^\infty (-1)^n\cdot\frac{2n}{n^2-1}\cdot\sin(nx)$ when $x\in [-\pi,\pi]$ To show this, I have used definitions for $\cos(x)$ and $\sin(x)$:
$$x\cdot \cos(x)+1/2\sin(x)=x\cdot \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\cdot x^{2n}+1/2\cdot \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\cdot x^{2n+1}$$
However, i do not how I'm supposed to proceed from here. You've got any ideas? This comes from the fourier series of $x\cdot \cos(x)$. We have the fourier coefficients as:
$$
c_{-1}=-\frac{i}4, c_0=0,c_1=\frac{i}4$$
and
$$c_n=(-1)^{n-1} \cdot \frac{in}{n^2-1}$$
when $|n|\geq 2$

Hint:
$$\dfrac{2n\sin nx}{(n+1)(n-1)}=\dfrac{\sin nx}{n-1}-\dfrac{\sin nx}{n+1}$$
Now $(-1)^n\dfrac{\sin nx}{n-1}$ is the imaginary part of $$\dfrac{(-1)^ne^{inx}}{n-1}=e^{ix}\cdot-\dfrac{(-e^{ix})^{n-1}}{n-1}$$
Now $$\sum_{n=2}^\infty-\dfrac{(-e^{ix})^{n-1}}{n-1}=\ln(1+e^{ix})=\ln (e^{ix/2})+\ln\left(2\cos\dfrac x2\right)=(2n\pi+\dfrac x2)i+\ln\left(2\cos\dfrac x2\right)$$
Put $n=0$ to find the principal value
Similarly for $$\dfrac{\sin nx}{n+1}$$
Finally  use How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
