This is problem 1.21 out of Lectures on the Fourier Transform and its Applications by Brad Osgood; we are using the definition of the Fourier series of a function $f(t)$ with period $T$ as
$$f(t) = \sum_{n=-\infty}^\infty c_ne^{\frac{2\pi int}{T}} \text{ where } c_n = \frac{1}{T} \int_0^T f(t)e^{\frac{-2\pi int}{T}}dt$$
In the previous part, I found the Fourier series of $f(t)=t^2$ on $t\in [0,2)$ periodized to have period $T=2$ to be
$$t^2= \frac{4}{3} + \sum_{n\neq0} \left( \frac{-2}{\pi in} + \frac{2}{\pi^2n^2} \right)e^{\pi int}$$
I've verified this with a few colleagues and am confident this is correct. I now need to use this result to solve the Basel problem and show $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$. I've been given the hint to let $t=0$, which gets close to the answer:
$$0^2= \frac{4}{3} + \sum_{n\neq0} \left( \frac{-2}{\pi in} + \frac{2}{\pi^2n^2} \right)e^{\pi in(0)}$$
$$0=\frac{4}{3} + \sum_{n\neq0} \frac{-2}{\pi in}\cdot 1 + \sum_{n\neq0} \frac{2}{\pi^2n^2}\cdot1$$
$$-\frac{4}{3} = \sum_{n=-\infty}^{-1}\frac{-2}{\pi in} + \sum_{n=1}^\infty \frac{-2}{\pi in} + \sum_{n=-\infty}^{-1} \frac{2}{\pi^2n^2} + \sum_{n=1}^\infty \frac{2}{\pi^2n^2}$$
$$-\frac{4}{3} = \sum_{n=1}^\infty \frac{-2}{\pi i(-n)} + \sum_{n=1}^\infty \frac{-2}{\pi in} + \sum_{n=1}^\infty \frac{2}{\pi^2(-n)^2} + \sum_{n=1}^\infty \frac{2}{\pi^2n^2}$$
$$-\frac{4}{3} = -\sum_{n=1}^\infty \frac{-2}{\pi in}+\sum_{n=1}^\infty \frac{-2}{\pi in} + \sum_{n=1}^\infty \frac{2}{\pi^2n^2} + \sum_{n=1}^\infty \frac{2}{\pi^2n^2}$$
$$-\frac{4}{3} = 0 + 2\sum_{n=1}^\infty \frac{2}{\pi^2n^2} = \frac{4}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2}$$
$$-\frac{\pi^2}{3}=\sum_{n=1}^\infty$$
But this is incorrect. I feel like saying $\sum_{n=1}^\infty \frac{-2}{\pi in} – \sum_{n=1}^\infty \frac{-2}{\pi in} = 0$ is wrong, because $\sum_{n=1}^\infty \frac{-2}{\pi in}$ diverges and $\infty-\infty$ is an indeterminant. After messing with it, I tried a hack by substituting $x=\sum_{n\neq0} \frac{-2}{\pi in}$ and got the result
$$0=\frac{4}{3} + x + \sum_{n\neq0} \frac{2}{\pi^2n^2} \implies
\frac{-(4+3x)}{2} \cdot \frac{\pi^2}{6} = \sum_{n=1}^\infty \frac{1}{n^2}$$
Setting $\frac{-(4+3x)}{2}=1$ gives
$$x=\sum_{n\neq0} \frac{-2}{\pi in}=-2$$
$$\sum_{n\neq0} \frac{1}{\pi in}=1$$
Which is telling me something is seriously wrong. I thought I was on the right track because the next part of the problem is to show $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}$. I can do this by setting $t=1$, assuming $\sum_{n\neq0} \frac{-2}{\pi in}=0$ and doing a similar calculation to what I did above. Does $\sum_{n\neq0} \frac{-2}{\pi in}=0$? If this is actually true, why? What is wrong with my calculation when $t=0$ and how can the hack of $x=-2$ be explained? If not, how can I handle the indeterminant? Is it possible to continue this approach with some modifications? Or is the only way to do this using Parseval's Theorem like in this similar problem?
Best Answer
Short version. Your computations are correct ... except the fact that $0^2 = 2$ in this context :-)
Detailed version. The famous Dirichlet convergence theorem for Fourier series tells that the Fourier series of a function $f$ periodic and piecewise $C^1$ converges to the mean-value of the left and right limits of the function, that is $$ \tilde{f}(t) = \sum_{n\in\Bbb Z} c_n(f)\,e^{i\pi nt} = \lim_{\varepsilon\to 0} \frac{f(t+\varepsilon)+f(t-\varepsilon)}{2} $$ In particular, the function defined by $f(t) = t^2$ on $[0,2)$ is discontinuous as a $2$-periodic function on $\Bbb R$, and $$ \tilde{f}(0) = \frac{2^2 + 0^2}{2} = 2. $$ This implies $$ 2 = \frac{4}{3} + \sum_{n\neq 0} \frac{2i}{\pi n} + \frac{2}{\pi^2 n^2}. $$ The $\frac{2i}{\pi n}$ term of the sum is indeed odd, and so vanishes (in the sense of writing the series as a limit of partial sums), and so the equation simplifies to $$ \frac{1}{3} = \sum_{n\neq 0} \frac{1}{\pi^2 n^2}, $$ which gives the famous $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}. $$