Prove $\sum_{i=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin{(2nx)}=\sum_{i=1}^{\infty}2\frac{(-1)^{n+1}}{n}\sin{(nx)}$

Question

I am required to prove the following equality:

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

For every $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Well, apparently, the Fourier Series of $f(x)=x$ is the RHS of the equation, which makes the equality easy to prove. The problem is that I wasn't aware of that. This question was a part of a Fourier Series exercise on the one hand, but on the other hand, I'm not sure whether am I required to know Fourier Series of some elementary functions by heart, or rather maybe there's a more sophisticated proof to the above equality.

Thank you very much!

Answer 1

$$-\sum_{n=1}^\infty\dfrac{(-1)^n\sin(2rnx)}n$$ is the imaginary part of $$-\sum_{n=1}^\infty\dfrac{(-1)^ne^{i2rnx}}n=-\sum_{n=1}^\infty\dfrac{(-e^{i2rx})^n}n=\ln(1-(-e^{i2rx}))$$

Now the principal value of $-\sum_{n=1}^\infty\dfrac{(-e^{i2rx})^n}n=\ln(1+e^{i2rx})$ is $$\ln(e^{irx})+\ln(e^{irx}+e^{-irx})=\underbrace{irx}_{\text{imaginary part}}+\ln(2\cos rx)$$

Here $r=\dfrac12, r=1$