the series: compute $ \sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^4} $ Compute
$$
\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^4}
$$
the result is $\frac{\pi^4+30\pi^2-384}{768}$, so I'm sure the sums $\sum\frac{1}{n^2}$ and $\sum\frac{1}{n^4}$ should appear in the solution.
The standard method $\frac{1}{4n^2-1}=\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})$ allows to compute $\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\frac{1}{2}$ and $\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^2}=\frac{\pi^2-8}{16}$, what about higher powers?
 A: Although it is tedious, you can use partial fractions to get: 
$\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(4n^2-1)^4}$ $= \dfrac{5}{32}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{2n+1}-\dfrac{1}{2n-1}\right)$ $+\dfrac{5}{32}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{(2n+1)^2}+\dfrac{1}{(2n-1)^2}\right)$ $+\dfrac{1}{8}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{(2n+1)^3}-\dfrac{1}{(2n-1)^3}\right)$ $+\dfrac{1}{16}\displaystyle\sum_{n = 1}^{\infty}\left(\dfrac{1}{(2n+1)^4}+\dfrac{1}{(2n-1)^4}\right)$. 
The first and third summations telescope to $-1$. 
Since $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \zeta(2) = \dfrac{\pi^2}{6}$, we have $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(2n)^2} = \sum_{n = 1}^{\infty}\dfrac{1}{4n^2} = \dfrac{1}{4}\zeta(2) = \dfrac{\pi^2}{24}$. 
Hence $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(2n-1)^2} = \dfrac{\pi^2}{6}-\dfrac{\pi^2}{24} = \dfrac{\pi^2}{8}$, and $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{(2n+1)^2} = \dfrac{\pi^2}{8} - 1$. 
You can do a similar thing for the fourth sum. 
