From $\frac{1}{2}+\sum_{n=1}^\infty \frac{4}{\pi^2 n^2}\cos(2\pi nx)=(1-2x)^2$ to $\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$

Question

I'm working on a proof of the following:

"Let $f(x)$ be the function defined by $f(x)=(1-2x)^2$ when $x\in [0,1)$ and extended to be $\mathbb{Z}$-periodic for the rest of the real line.

(a) Show that the series $\frac{1}{2}+\sum_{n=1}^\infty \frac{4}{\pi^2 n^2}\cos(2\pi nx)$ converges uniformly to $f$;

(b) Conclude that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$;

(c) Conclude that $\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$."

Now, I managed to prove $(a)$ and $(b)$ but I'm stuck on $(c)$ so I'd appreciate any hint about how to finish this last step.

(Note: I know there are other questions on this topic but they use different series, so this is not a duplicate)

Answer 1

(c) $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{4}{\pi^2n^2}\cos(2\pi nx) = (2x-1)^2-\frac{1}{3} $

=> $\enspace\displaystyle \int\limits_0^x\sum\limits_{n=1}^\infty \frac{1}{\pi^2n^2}\cos(2\pi nt)dt = \int\limits_0^x ((t-\frac{1}{2})^2-\frac{1}{12})dt $

=> $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{1}{\pi^3n^3}\sin(2\pi nx) = 2\frac{(x-\frac{1}{2})^3}{3}+\frac{1}{12}-\frac{x}{6} $

=> $\enspace\displaystyle \int\limits_0^x\sum\limits_{n=1}^\infty \frac{1}{\pi^3n^3}\sin(2\pi nt) dt = \int\limits_0^x (2\frac{(t-\frac{1}{2})^3}{3}+\frac{1}{12}-\frac{t}{6} )dt $

=> $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{1}{\pi^4n^4}(1-\cos(2\pi nx)) = \frac{(x-\frac{1}{2})^4}{3}-\frac{1}{48}+\frac{x}{6}-\frac{x^2}{6} $

=> $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{1}{\pi^4n^4}(1-(-1)^n) = \frac{1}{48} $

=> $\enspace\displaystyle \sum\limits_{n=0}^\infty \frac{1}{(2n+1)^4} = \frac{\pi^4}{96} $

=> $\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{1}{n^4} = \frac{1}{1-\frac{1}{2^4}}\frac{\pi^4}{96} = \frac{\pi^4}{90} $

Answer 2

Hint: use Parseval's theorem to relate the series of the squared coefficients to $\int_0^1 f(x)^2dx$.