# 2016 UNCO Math Contest II Problems/Problem 7

## Problem

Evaluate $$S =\sum_{n=2}^{\infty} \frac{4n}{(n^2-1)^2}$$

## Solution

First, we do fraction decomp. Let $$\frac{A}{(n-1)^2}+\frac{B}{(n+1)^2} = \frac{4n}{(n^2-1)^2}$$. Multiplying both sides by $(n^2-1)^2$ and expanding gives $$(A+B)n^2+2(A-B)n+(A+B)=4n$$ Therefore, we have the system of equations $$\begin{cases} A+B=0\\ A-B=2\end{cases}$$. Adding the two equations gives $2A=2 \implies A=1$, while subtracting the two gives $2B=-2 \implies B=-1$.

Therefore, $$\frac{4n}{(n^2-1)^2}=\frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}$$, so $$S =\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}$$ $$= \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} - \sum_{n=2}^{\infty} \frac{1}{(n+1)^2}$$

Writing out the first few terms and rearranging, we have $$\frac{1}{1^2}+\frac{1}{2^2}+\left(\cancel{\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)- \left(\cancel{\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)$$, which telescopes to $\frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}$ -NamelyOrange

## Solution 2

This is a telescoping series:

(1−1/9)+(1/4−1/16)+(1/9−1/25)+(1/16−1/36)+(1/25−1/49)+...=5/4