Infinite sums and products 3 (11)

We can write the sum as $\displaystyle{\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2(2n+1)^2}}$ .Applying partial fraction decomposition gives us: $\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2(2n+1)^2}=\frac{1}{4}\sum^{\infty}_{n=1}\left(\frac{1}{2n+1}+\frac{1}{(2n+1)^2}-\frac{1}{2n-1}+\frac{1}{(2n-1)^2}\right)$ $=\frac{1}{4}\sum^{\infty}_{n=1}\left(\frac{1}{2n+1}-\frac{1}{2n-1}\right)+\frac{1}{4}\sum^{\infty}_{n=1}\left(\frac{1}{(2n+1)^2}+\frac{1}{(2n-1)^2}\right)$ Now we will deal with each summation separately.First deal with the first summation,which is a telescoping sum.Writing out the first few terms of the series,we see that all terms but $\frac{-1}{1}$ get cancelled out.So: $\frac{1}{4}\sum^{\infty}_{n=1}\left(\frac{1}{2n+1}-\frac{1}{2n-1}\right)=\frac{1}{4}\times\frac{-1}{1}=-\frac{1}{4}$ The second sum is a little bit complicated.Writing out the first few terms,we get: $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{5^2}+\dots\\=1+2\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots\right)\\1+2\left(\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots\right)-1-\left(\frac{1}{2^2}+\frac{1}{4^2}+\dots\right)\right)\\=1+2\left(\frac{\pi^2}{6}-1-\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\dots\right)\right)\\=1+2\left(\frac{\pi^2}{6}-1-\frac{1}{4}\times\frac{\pi^2}{6}\right)\\=1+2\left(\frac{\pi^2}{6}-1-\frac{\pi^2}{24}\right)\\=1+2\left(\frac{\pi^2}{8}-1\right)\\=1+\frac{\pi^2}{4}-2=\frac{\pi^2}{4}-1$ So the summation equals $\frac{1}{4}\times\left(\frac{\pi^2}{4}-1\right)=\frac{\pi^2}{16}-\frac{1}{4}$ So the series in the question equals $\frac{-1}{4}+\frac{\pi^2}{16}-\frac{1}{4}=\frac{\pi^2}{16}-\frac{1}{2}$