$\displaystyle\sum_{n=1}^{\infty}\dfrac{4n}{(4n^2-1)^2}=$
Step-by-Step Solution
Key Concept: Partial fractions: $\frac{4n}{(4n^2-1)^2}=\frac{4n}{(2n-1)^2(2n+1)^2}$
$\frac{4n}{(2n-1)^2(2n+1)^2}=\frac{1}{4}\left(\frac{1}{(2n-1)^2}-\frac{1}{(2n+1)^2}\right)\cdot\frac{1}{2}$... After telescoping: sum $=\frac{1}{4}$.
Correct Answer: 3