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 $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}$,using problem 5.8.

Short answer

By Parseval theorem$\frac{{\pi }^2}{6}=\sum _{n=1}^{\infty }\frac{1}{n^2}$

Step-by-step solution

Given Information.

The given series is$\sum _{n=1}^{\infty }\frac{1}{n^2}$.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 5.8 is$f\left(x\right)=1-2\sum _{n=1}^{\infty }\frac{\sin \left(nx\right)}{n}{\left(-1\right)}^n$

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

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

Use the Parseval theorem

$$\begin{gathered} 1+\frac{{\pi }^2}{3}=1+\frac{1}{2}\sum _{n=1}^{\infty }{\left(\frac{2{\left(-1\right)}^n}{n}\right)}^2 \\ \Rightarrow \frac{{\pi }^2}{6}=\sum _{n=1}^{\infty }\frac{1}{n^2} \end{gathered}$$

Thus $\frac{{\pi }^2}{6}=\sum _{n=1}^{\infty }\frac{1}{n^2}$