Question

Use Parseval’s Theorem and the results of the indicated problems to find the sum of the series in Problems 5to 9

The series $\frac{\mathbf{1}}{{\mathbf{3}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{15}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{35}}^{\mathbf{2}}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}$, using Problem 5.11

Short answer

The value of the series$\frac{1}{3^2}+\frac{1}{{15}^2}+\frac{1}{{35}^2}+...$ is$\frac{{\pi }^2}{16}-\frac{1}{2}$ by the use of Parseval’s theorem

Step-by-step solution

Definition of Parseval’s Theorem

In mathematics, Parseval's theorem generally refers to the result that the Fourier transfigure is unitary; approximately, that the sum of the forecourt of a function is equal to the sum of the forecourt of it is transfigure.

 Step 2: Given Parameters

Given the series$\frac{1}{3^2}+\frac{1}{{15}^2}+\frac{1}{{35}^2}+...$

Value of the series is to be found using Parseval Theorem and Problem 5.11.

Use the function from problem 5.11

From Problem 5.11, write the function as$f\left(x\right)=\frac{1}{\pi }+\frac{1}{2}\sin x-\frac{2}{\pi }\sum _{n-even}^{\infty }\frac{\cos \left(nx\right)}{n^2-1}$where$n\ne 0$

Find the mean value of square of the function

Now, use the Parseval Theorem to get

$\frac{1}{4}=\frac{1}{{\pi }^2}+\frac{1}{2}{\left(\frac{1}{2}\right)}^2-\frac{1}{2}\sum _{n-even}^{\infty }{\left(\frac{2}{\pi \left(n^2-1\right)}\right)}^2$

Solve further to get

$\sum _{n-even}^{\infty }\frac{1}{{\left(n^2-1\right)}^2}=\frac{{\pi }^2}{16}-\frac{1}{2}$

Therefore, the sum of series$\frac{1}{3^2}+\frac{1}{{15}^2}+\frac{1}{{35}^2}+...$ is$\frac{{\pi }^2}{16}-\frac{1}{2}$ by the use of Parseval’s theorem.