I have been solving problems using a "Potentially Helpful Formulas" sheet from my esteemed math professor. i want to solve for:
$\sum_{n=1}^{\infty} \dfrac{1}{n^4} =$ ?
On my formula sheet i have:
$\sum_{n=1}^{\infty} \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$
Which looks similar to what i want, but not quite. Might someone guide me towards the solution (or provide the solution with explanation?
This is just a sketch of one of many possible proofs.
Step1. Prove that over the interval $[0,2\pi]$, the function: $$f(x)=\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n^2}$$ is a second degree-polynomial whose graphics passes through the points: $$(0,\pi^2/6),\quad (\pi,-\pi^2/12),\quad (2\pi,\pi^2/6).$$
Step2. Deduce from Lagrange interpolation that: $$ f(x) = \frac{\pi^2}{6}-\frac{x(2\pi-x)}{4}.$$
Step3. Apply Parseval's identity to $f(x)$: $$\int_{0}^{2\pi}f(x)^2\,dx = \pi\sum_{n=1}^{+\infty}\frac{1}{n^4}.$$
Step4. Prove, through the second step, that: $$\int_{0}^{2\pi}f(x)^2\, dx = \frac{\pi^5}{90}.$$
Conclusion:
$$\zeta(4)=\sum_{n=1}^{+\infty}\frac{1}{n^4} = \frac{\pi^4}{90}.$$
