So...I previously found the series for $ f(t)= \begin{cases} 0&\text{if}\, -\pi\leq t\lt -\pi/2\\ \cos(t)&\text{if}\, -\pi/2\leq t\leq \pi/2\\ 0&\text{if}\, \pi/2\lt t\leq \pi\\ \end{cases} $

Where its period,$T$, is $2\pi$ and thus $w=2\pi/T=1$, and its series is:


I should be able to "easily" find the sum of the series of $\sum_{n=1}^∞\frac{(-1)^{n+1}}{4n^2-1}$ by using Parseval's identity exclusively

I know, by Parseval, and after some simplification, that $\sum_{n=2}^∞\frac{-2\cos(\frac{n\pi}{2})}{\pi(n^2-1)}\cos(nt)$=$\frac{\pi-2}{2\pi}$

But I am struggling at actually simplifying $\sum_{n=1}^∞\frac{(-1)^{n+1}}{4n^2-1}$ to the point that I am thinking that something here is wrong. So far, I have:


I am not sure if/how I should be evaluating $t$ in my original series, maybe that way I can make $-\cos(n\pi)=\cos(\frac{n\pi}{2})cos(nt)$ and afterwards somehow simplify the divisor?

Any suggestions/corrections are welcome

Edit 1:

By evaluating $t=\pi/2$ on the original series, I have:


Which is somewhat promising, though I haven't been able to far to make it similar to the desired series. I have tried messing with partial fractions and checking for patterns on how they evaluate for the first $n$'s, but no luck so far.

  • $\begingroup$ I think you should set $t = \frac{\pi}{2}$ and use double angle formulae to get $\cos(n \pi)$. $\endgroup$
    – fGDu94
    May 13, 2019 at 10:39
  • $\begingroup$ Thanks! Indeed that gets it closer, though no luck so far, I added more info on the Edit. $\endgroup$
    – Lightsong
    May 15, 2019 at 8:44

1 Answer 1


You have shown that the series for $f(t)$ is $$ f(t)\sim \frac{1}{2}+\frac{1}{\pi}+\sum_{n=2}^{\infty}\frac{-2\cos\frac{n\pi}{2}}{\pi(n^2-1)}\cos(nt). $$ The terms of the sum are $0$ for $n=3,5,7,\cdots$. So, $f$ may be written as $$ f(t) \sim \frac{1}{2}+\frac{1}{\pi}+\sum_{k=1}^{\infty}\frac{-2\cos(k\pi)}{\pi(4k^2-1)}\cos(2kt) \\ \;\;\;\;\;\;= \frac{1}{2}+\frac{1}{\pi}-2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{\pi(4k^2-1)}\cos(2kt) $$ The function $f$ is continuous and differentiable at $t=0$. So $f(0)$ is equal to the Fourier series. $f(0)=1$ gives $$ 1 = \frac{1}{2}+\frac{1}{\pi}-2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{\pi(4k^2-1)} \\ \frac{\pi}{2}\left(\frac{1}{\pi}-\frac{1}{2}\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{4k^2-1} $$

  • $\begingroup$ Wow, thanks! That was quite tricky to get. Why did you square $\pi$ in the second step on the series denominator though? $\endgroup$
    – Lightsong
    May 17, 2019 at 8:49
  • $\begingroup$ @Lightsong : That looks like a mistake. Suddenly just replaced $\pi$ with $\pi^2$. So I just changed it. The tricky part was making sure to get the Fourier series right, which is what you already did. $\endgroup$ May 17, 2019 at 11:20

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