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.
Best Answer
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 $$ 1 = \frac{1}{2}+\frac{1}{\pi}-2\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{\pi(4k^2-1)} \\ \frac{\pi}{2}\left(\frac{1}{\pi}-\frac{1}{2}\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{4k^2-1} $$