Question

Use Problem 5.7to show that$\frac{\mathbf{1}}{{\mathbf{2}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{4}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{6}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{\cdots }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Short answer

The resultant expansion is $\sum _{n=2k}^{\infty }\frac{1}{n^2-1}=\frac{1}{2}k=1,2,3,4$.

Step-by-step solution

Given data

The given function is $\frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+\cdots =\frac{1}{2}$.

Concept of Dirichlet’s theorem

Dirichlet's theorem states that the collection of all numbers of the form$\mathit{n}\mathit{a}\mathbf{+}\mathit{b}$contains an unlimited number of prime numbers, where the constants a and b are integers with no common divisors other than 1 and the variable nis any natural number.

Simplify the expression

Use the Problem (5.11) with $x=0$.

$$\begin{gathered} f\left(0\right)=0 \\ f\left(0\right)=\frac{1}{\pi }+0-\frac{2}{\pi }\sum _{n=2k}^{\infty }\frac{1}{n^2-1} \\ f\left(0\right)=0\sum _{n=2k}^{\infty }\frac{1}{n^2-1} \\ f\left(0\right)=\frac{1}{2} \end{gathered}$$

Therefore, expansion is $\sum _{n=2k}^{\infty }\frac{1}{n^2-1}=\frac{1}{2}k=1,2,3,4$.