For the series $\displaystyle\sum_{n=0}^{\infty}\frac{1}{((2n+1)^2-4)^2}$ wolframalpha gives answer $\displaystyle\frac{\pi^2}{64}$, but another site gives $\displaystyle\frac{\pi^2}{64}-\frac{1}{12}$: http://www.emathhelp.net/calculators/calculus-2/series-calculator/?f=1%2F%28%282n%2B1%29%5E2-4%29%5E2&var=&a=0&b=inf.
What is the correct answer?
$$\begin{align} S&=\sum_{n=0}^\infty\frac{1}{((2n+1)^2-4)^2}=\sum_{n=0}^\infty\frac{1}{(4n^2+4n-3)^2}=\sum_{n=0}^\infty\frac{1}{((2n-1)(2n+3))^2}\\ &=\frac{1}{4}\sum_{n=0}^\infty\frac{2n+3-(2n-1)}{((2n-1)(2n+3))^2}=\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)^2(2n+3)}-\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)(2n+3)^2}\\ &=\frac{1}{16}\sum_{n=0}^\infty\frac{1}{(2n-1)^2}-\frac{2}{16}\sum_{n=0}^\infty\frac{1}{(2n-1)(2n+3)}+\frac{1}{16}\sum_{n=0}^\infty\frac{1}{(2n+3)^2} \end{align}$$ Using this we see $$\begin{align} &\sum_{n=0}^\infty\frac{1}{(2n-1)^2}=\frac{3}{4}\zeta(2)+1\\ &\sum_{n=0}^\infty\frac{1}{(2n+3)^2}=\frac{3}{4}\zeta(2)-1 \end{align}$$ And we can use telescoping on the middle-term as follows $$\begin{align} \sum_{n=0}^\infty\frac{1}{(2n-1)(2n+3)}&=\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)}-\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n+3)}\\ &=\frac{1}{4}\sum_{n=0}^\infty\frac{1}{(2n-1)}-\frac{1}{4}\sum_{n=2}^\infty\frac{1}{(2n-1)}\\ &=\frac{1}{4}(-1+1)=0 \end{align}$$ And so $$S=\frac{1}{16}\times\frac{3}{2}\zeta(2)=\frac{\pi^2}{64}$$
