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!

https://math.stackexchange.com/questions/3361589/proof-that-sum-limits-n-1-infty-1n1-sinn-overn-1-over2 — lab bhattacharjee, Nov 13 '19 at 07:46
@labbhattacharjee It's more complicated here since there $n=1$, but here $n\in\mathbb Z$ — Amit Zach, Nov 13 '19 at 07:51
I'm not sure if this is true, because at $x=\frac{\pi}{2}$, the LHS is $0$, but the RHS becomes $$\sum_{k=0}^\infty 2\frac{(-1)^k}{2k+1}=2\arctan(1) = \frac{\pi}{2}$$ — Ninad Munshi, Nov 13 '19 at 08:18

score 2 · Answer 1 · answered Nov 13 '19 at 08:29

2

$$-\sum_{n=1}^\infty\dfrac{(-1)^n\sin(2rnx)}n$$ 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 $r=\dfrac12, r=1$

answered Nov 13 '19 at 08:29

lab bhattacharjee

274,582

Thank you very much! – Amit Zach Nov 13 '19 at 09:40

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

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