Nevermind, I just forgot what cosine is at pi/2...
... Well, even though I solved that last part, I'm now stuck trying to find the value of $\displaystyle \displaystyle \sum^{\infty}_{n=1} \frac{1}{(2n-1)^{2}}$. I kind of suspect I should be able to get this from Parseval's Theorem, given in Apostol as $\displaystyle \frac{1}{\pi} \displaystyle \int^{2 \pi}_{0} |f(x)|^{2} dx = \frac{a_{0}^{2}}{2} + \sum^{\infty}_{n=1} (a_{n}^{2} + b_{n}^{2})$. I suspect this because, somehow, the teacher claimed to have used this to show that $\displaystyle \displaystyle \sum^{\infty}_{n=1} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$.
Anyway, I plugged in the constant function $\displaystyle \frac{\pi}{2}$ and got a truism, so in general I can tell I don't want to use constant functions. I suppose I want to find some function such that $\displaystyle a_{n}^{2} + b_{n}^{2}$ is equal to the sum I'm looking to evaluate, but I can't see how to make this magically happen.
Notice that
$\displaystyle \frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\fr ac{1}{3^2}+\dots\right) + \left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\d ots\right)$
$\displaystyle = \frac{1}{1^2}+\frac{1}{2^2}+\dots =\frac{\pi^2}{6}$,
from which you can just solve. (Note that all manipulations are justified by absolute convergence of the series.)
I'm afraid I don't follow. If this is a proof that $\displaystyle \displaystyle \sum^{\infty}_{n=1} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ I'm not exactly sure what is taken as known, from which we prove this. However, here is my (somewhat failed) attempt at reproducing the professor's argument: Using Parseval, $\displaystyle \frac{\pi^{2}}{3} = \frac{a_{0}}{2} + \displaystyle \sum^{\infty}_{n=1}(a_{n}^{2} + b_{n}^{2})$ where $\displaystyle a_{0} = 2\pi$ and $\displaystyle a_{n} = 0$ and $\displaystyle b_{n} = \frac{2}{n}$ so $\displaystyle \frac{\pi^{2}}{3} = 2\pi + \displaystyle 4\sum^{\infty}_{n=1} \frac{1}{n^{2}} \Rightarrow$ I get the wrong answer.
In any case, I've used another proof of the same fact, and I'm still not sure how to solve the problem I'm faced with.