How to prove this equation with fourier transformation?

Question

I confronted with the following equation

$\sum_{1}^{\infty}(-1)^{n-1}\frac{sin(\omega n)}{n}=\frac{\omega}{2}$

I guess fourier transformation is needed, but it's too complicated to work out. Hope someone could help.

Answer 1

If you are trying to solve this for $\omega$, then notice that the function $f(x)=\frac{x}{2}$, $-\pi \leq x \leq \pi$, $f(x)=f(x+2\pi)$ has the fourier series representation $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}sin(nx)$, which is explained fully at https://en.wikipedia.org/wiki/Fourier_series#Example_1:_a_simple_Fourier_series, so setting $x=\omega$, you have the equality $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}sin(n\omega)=\frac{\omega}{2}$, $-\pi \leq \omega \leq \pi$,

Answer 2

We have geometric series: $$ \dfrac{1}{1+x}=1-x+x^2-x^3+...=\sum_{n=0}^{\infty}(-1)^nx^n.$$ Take integral of that; integration term by term (constant of integration is zero): $$ln(1+x) =\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{n+1}}{n+1} = \sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^{n}}{n}.$$ Let $x=e^{ix}$. We have: $$ ln(1+e^{ix})= \sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{e^{inx}}{n}=\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{cos(nx)+isin(nx)}{n}.$$ Imaginary part on the right must be imaginary part on the left: $$\Im\{ln(1+e^{ix})\}=\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{sin(nx)}{n}. $$ But: $$\Im\{ln(1+e^{ix})\} = \dfrac{x}{2} $$ Q.E.D.