Question

Use Parseval’s theorem and the results of the indicated problems to find the sum of the series in Problems 5 to 9. The series $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{o}\mathbf{d}\mathbf{d}}^{}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{4}}}$ ,using problem 9.10.

Short answer

By Parseval theorem$\sum _{n=odd}^{\infty }\frac{1}{n^4}=\frac{{\pi }^4}{96}$

Step-by-step solution

Given Information.

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

Definition of Parseval’s theorem

According to Parseval's Theorem, a signal's energy can be defined as the average energy of its frequency components.

Sum of the series

It is know that the solution of the problem 9.10is $f\left(x\right)=\frac{\pi }{4}-\frac{2}{\pi }\sum _{n=odd}^{\infty }\frac{\cos \left(2nx\right)}{n^2}$

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

$<{\left|f\left(x\right)\right|}^2>=\frac{1}{\pi }\underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{x}^{2}dx=\frac{1}{3\pi }{x}^3{{|}^{\frac{\pi }{2}}}_{\frac{-\pi }{2}}=\frac{{\pi }^2}{12}$

Use the Parseval theorem

$$\begin{gathered} \frac{{\pi }^2}{12}=\frac{{\pi }^2}{16}+\frac{1}{2}\sum _{n=odd}^{\infty }\left(\frac{4}{{\pi }^2{n}^4}\right) \\ \Rightarrow \frac{2}{{\pi }^2}\sum _{n=odd}^{\infty }\frac{1}{n^4}=\frac{{\pi }^2}{48} \\ \Rightarrow \sum _{n=odd}^{\infty }\frac{1}{n^4}=\frac{{\pi }^2}{96} \end{gathered}$$

Thus $\sum _{n=odd}^{\infty }\frac{1}{n^4}=\frac{{\pi }^4}{96}$