How to find sum and convergence of series $\sum_{n = 1}^{\infty}\frac{1}{(2n-1)^{2}\cdot(2n+1)^{2}}$

Question

I laid this shot that: $$\frac{1}{(2n-1)^2(2n+1)^2}=1/4\, \left( 2\,n+1 \right) ^{-2}-1/4\, \left( 2\,n-1 \right) ^{-1}+1/ 4\, \left( 2\,n-1 \right) ^{-2}+1/4\, \left( 2\,n+1 \right) ^{-1} $$ But I don't know what to do next. Help me please

Answer 1

Note that:

$$\sum_{n=1}^\infty\frac1{(2n-1)^2}=1+\sum_{n=1}^\infty\frac1{(2n+1)^2}$$

$$\sum_{n=1}^\infty\frac1{2n-1}=1+\sum_{n=1}^\infty\frac1{2n+1}$$

Thus, your series reduces as follows:

$$\sum_{n=1}^\infty\frac1{(2n-1)^2(2n+1)^2}=-\frac12+\frac12\sum_{n=1}^\infty\frac1{(2n-1)^2}$$

Since everything cancels except that. Now let $S$ be the following sum:

$$S=\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$$

It can be seen that

$$S=\sum_{n=1}^\infty\left[\frac1{(2n-1)^2}+\frac1{(2n)^2}\right]=\frac14S+\sum_{n=1}^\infty\frac1{(2n-1)^2}$$

Thus,

$$\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac34S=\frac{\pi^2}8$$

and finally,

$$\sum_{n=1}^\infty\frac1{(2n-1)^2(2n+1)^2}=-\frac12+\frac12\left(\frac{\pi^2}8\right)=\frac{\pi^2-8}{16}$$

Answer 2

By partial fraction decomposition $$ \frac{1}{(2x-1)^2 (2x+1)^2} = \frac{\frac{1}{16}}{\left(x-\frac{1}{2}\right)^2}+\frac{\frac{1}{8}}{x-\frac{1}{2}}-\frac{\frac{1}{8}}{x+\frac{1}{2}}+\frac{\frac{1}{16}}{\left(x+\frac{1}{2}\right)^2} \tag{1}$$ hence it follows that: $$ \sum_{n\geq 1}\frac{1}{(2n-1)^2 (2n+1)^2}=\frac{1}{4}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n-1)^2}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n+1)^2} = \color{red}{\frac{\pi^2-8}{16}}\tag{2}$$