Evaluate infinite sum using Parseval's theorem (Fourier series)

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SUMMARY

The infinite sum \(\sum_{n=1}^{\infty}\frac{1}{n^4}\) equals \(\frac{\pi^4}{90}\) as demonstrated using Parseval's theorem. The theorem states that \(\frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2dx = \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)\). By applying this theorem to the function \(f(x) = x^2\) over the interval \([-π, π]\), the solution can be derived effectively. This approach confirms the relationship between Fourier series coefficients and the evaluated infinite sum.

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  • Knowledge of integration techniques over defined intervals
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Homework Statement


Show that: \sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{π^4}{90}
Hint: Use Parseval's theorem

Homework Equations


Parseval's theorem:

\frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2dx = \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)

The Attempt at a Solution


I've been trying to solve this for ages and I just can't figure out what to do. I know you're supposed to use Parseval's theorem. All I've managed to do was plug in \frac{1}{n^4} into the summation part of the Parseval's equation and I substituted the formula for a0 but I couldn't get very far.

Any help would be really appreciated.
 
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Try the function f(x)=x^2 for x\in [-\pi,\pi].
 
It works, ty
 

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