# Finding the sum of another Fourier series using Parseval's identity

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:

$$\frac{1}{2}+\frac{1}{\pi}+\sum_{n=2}^∞\frac{-2\cos(\frac{n\pi}{2})}{\pi(n^2-1)}\cos(nt)$$

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:

$$\sum_{n=1}^∞\frac{(-1)^{n+1}}{4n^2-1}=\frac{1}{3}+\sum_{n=2}^∞\frac{(-1)^{n+1}}{4n^2-1}=\frac{1}{3}+\sum_{n=2}^∞\frac{-\cos(n\pi)}{4n^2-1}$$

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:

$$\frac{1}{2}+\frac{1}{\pi}+\sum_{n=2}^∞\frac{\cos(n\pi)+1}{\pi(n^2-1)}$$

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.

• I think you should set $t = \frac{\pi}{2}$ and use double angle formulae to get $\cos(n \pi)$. – fGDu94 May 13 '19 at 10:39
• Thanks! Indeed that gets it closer, though no luck so far, I added more info on the Edit. – Lightsong May 15 '19 at 8:44

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 $$1 = \frac{1}{2}+\frac{1}{\pi}-2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{\pi(4k^2-1)} \\ \frac{\pi}{2}\left(\frac{1}{\pi}-\frac{1}{2}\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{4k^2-1}$$
• Wow, thanks! That was quite tricky to get. Why did you square $\pi$ in the second step on the series denominator though? – Lightsong May 17 '19 at 8:49
• @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. – COVID-20 May 17 '19 at 11:20