# How to prove this equation with fourier transformation?

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.

• What do you mean "equation" here, and what do you need to do? – DonAntonio Nov 8 '18 at 0:05
• Just looking on Wikipedia en.wikipedia.org/wiki/…, it seems like your sum is the fourier series representation for a sawtooth wave $y=\frac{x}{2}$, $-\pi \leq x \leq \pi$, $y(x)=y(x+2\pi)$ – Seth Nov 8 '18 at 0:07
• If you mean to solve for $\;\omega\;$ , take $\;\omega=0\;$ for example, and you get $$\color{red}{\sum_{k=1}^\infty(-1)^{n-1}\frac{\sin(0\cdot n)}n}=0=\color{red}{\frac02}$$ – DonAntonio Nov 8 '18 at 0:08
• I think that any solution $-\pi \leq \omega \leq \pi$ would work, if the wiki article I read is correct – Seth Nov 8 '18 at 0:09

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 $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}sin(n\omega)=\frac{\omega}{2}$$, $$-\pi \leq \omega \leq \pi$$,
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.