Evaluate the sum of series convergent series Compute     $\sum\frac{1}{n^2(2n-1)}$ assuming that $\sum\frac{1}{n^2}$ = $\frac{\pi^2}{6}$
If we multiply and divide the fraction by 2n+1 we get $\sum\frac{2n+1}{n^2(4n^2-1)}$
but still I couldn't proceed.  How to use $\sum\frac{1}{n^2}$ = $\frac{\pi^2}{6}$
to evaluate the series sum?

\begin{align*}
\sum\dfrac{1}{n^{2}(2n-1)}&=-\sum\dfrac{2n-1-2n}{n^{2}(2n-1)}\\
&=-\sum\dfrac{1}{n^{2}}+2\dfrac{1}{n(2n-1)}\\
&=-\sum\dfrac{1}{n^{2}}+4\sum\left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right)
\end{align*}
