Showing that the limit of $\sum_{k=0}^{\infty}\frac{1}{(k+\frac{1}{2})^2} = \frac{\pi^2}{2}$ by taking a look at the result of a Fourier series

Question

What I found out: If we turn the function $$f(x):=\begin{cases} \, 1 & 0 < x < \pi \\ 0 & x = 0 \\ -1 & -\pi < x < 0\end{cases}$$ into a Fourier series, we get the result (evaluated at $x = \pi/2$, I justify this by the pointwise convergence of the Fourier series at values $x >0$ to $1$) $$\sum_{n=0}^{\infty}\frac{(-1)^n}{n+\frac{1}{2}} = \frac{\pi}{2} \enspace$$ This comes from the (simplified) Fourier series of f: $$\sum_{k=0}^{\infty}\frac{2\sin{[(2n+1)x]}}{(n+\frac{1}{2})\pi}$$ I found this problem: Using the result you find by turning f into its Fourier series, find the value of $$\sum_{i=0}^{\infty}\frac{1}{(i+\frac{1}{2})^2}$$ Spoiler Alert: using Maple I found that the above sum converges to $\frac{\pi^2}{2}$. I am stuck on how to proceed, I tried to take the Cauchy product of the series with itself, only to notice I can't even do that because the series above is not absolutely convergent. I think I'm missing just one step, but I can't seem to find it. (Might I have to modify the function f above to say $x$ instead of $1$? I'll try that now but I don't think that is it)

Answer 1

HINT:

If $$f(x)=\sum_{n=0}^\infty \frac{2\sin((2n+1)x)}{\pi(n+1/2)}$$ then $$\int f(x)dx+C=-\sum_{n=0}^\infty \frac{2\cos((2n+1)x)}{\pi(n+1/2)(2n+1)}$$

Answer 2

You can use Parseval's theorem. This states that if $$ f(x) = \sum_{n=1}^\infty b_n \sin(nx)$$ then $$ \int_{-\pi}^\pi f^2 dx = \pi \sum_{n=1}^\infty b_n^2.$$ Since you already know the $b_n$'s, all that remains is for you to evaluate the integral of $f^2$, and plug the $b_n$'s into the formula.