prove that
$\displaystyle \frac{1}{1^2.3^3.5^2}-\frac{1}{3^2.5^3.7^2}+\frac{1}{5^2.7^3.9^2}-........=\frac{1}{9}-\frac{\pi}{2^6}-\frac{\pi^3}{2^9}$
Some things come to mind:
The sum is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)^2 (2n+3)^3 (2n+5)^2} = \sum_{n=0}^\infty (-1)^{n} \left( \frac{A}{(2n+1)^2} + \frac{B}{(2n+3)^3} + \frac{C}{(2n+5)^2} \right) $ where A, B, C are easily found (but not so easily that I can be bothered doing it).
Then break this up into three sums. Some re-labelling of the dummy index might be necessary after doing the following:
Have a look here for some series that might be useful .....
I'm too lazy to work through the details, so be warned - what I've suggested might not lead anywhere .......