2016 UNCO Math Contest II Problems/Problem 7

Problem

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

Solution

First, we perform fractional decomposition on the summed expression.

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

$\boxed{\frac{5}{4}}$

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

2016 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions

Page

Toolbox

Search

2016 UNCO Math Contest II Problems/Problem 7

Contents

Problem

Solution

Solution 2

See also