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 $\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{3}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{5}}^{\mathbf{2}}}\mathbf{+}\mathbf{\cdots }\mathbf{,}$ using problem 9.6.

Short answer

By Parseval theorem$\frac{{\pi }^2}{8}=\sum _{n=odd}^{\infty }\frac{1}{n^2}$

Step-by-step solution

Given Information.

The given series is$1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots$.The sum of the series is to be found out.

Definition of Parseval’s theorem& average value

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

The average value of the square of the function over the interval is 1

Sum of the series

It is know that the solution of the problem 9.6 is $f\left(x\right)=\frac{4}{\pi }\sum _{n=odd}^{\infty }\frac{sin\left(\frac{\pi nx}{l}\right)}{n}$

Use the Parseval theorem and the fact that the average value of the square of the function over the interval is 1

Now, $$\begin{gathered} <{\left|f\left(x\right)\right|}^2>=1=\sum _{n=odd}^{\infty }\frac{1}{2}{\left(\frac{4}{\pi n}\right)}^2 \\ \Rightarrow \frac{{\pi }^2}{8}=\sum _{n=odd}^{\infty }\frac{1}{n^2} \end{gathered}$$

Thus $\frac{{\pi }^2}{8}=\sum _{n=odd}^{\infty }\frac{1}{n^2}$