Question

Use Parseval’s theorem and the results of the indicated problems to find the sum of the series in Probllems 5 to 9. The series $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{4}}}$ ,using problem 9.9.

Short answer

By Parseval theorem$\frac{{\pi }^4}{90}=\sum _{n=1}^{\infty }\frac{1}{n^4}$

Step-by-step solution

Given Information.

The given series is$\sum _{n=1}^{\infty }\frac{1}{n^4}$.The sum of the series is to be found out.

Definition of Parseval’s theorem

Parseval’s theorem is a theorem stating that the energy of a signal can be expressed as its frequency components’ average energy.

Sum of the series

It is know that the solution of the problem 9.9 is$f\left(x\right)=\frac{1}{12}+\frac{1}{{\pi }^2}\sum _{n=1}^{\infty }\frac{\cos \left(2\pi nx\right)}{n^2}{\left(-1\right)}^n$

It is known that the Mean value of the square of the function is

$<{\left|f\left(x\right)\right|}^2>=\underset{-\frac{1}{2}}{\overset{\frac{1}{2}}{\int }}{x}^{4}dx=\frac{1}{5}{x}^5{{|}_{\frac{-1}{2}}}^{\frac{1}{2}}=\frac{1}{80}$

Use the Parseval theorem

$$\begin{gathered} \frac{1}{80}=\frac{1}{144}+\frac{1}{2}\sum _{n=1}^{\infty }{\left(\frac{{\left(-1\right)}^n}{{\pi }^2{n}^2}\right)}^2 \\ \Rightarrow \frac{{\pi }^4}{90}=\sum _{n=1}^{\infty }\frac{1}{n^4} \end{gathered}$$

Thus $\frac{{\pi }^4}{90}=\sum _{n=1}^{\infty }\frac{1}{n^4}$