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

2016 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2016 UNCO Math Contest II Problems/Problem 7

Contents

Problem

Solution

Solution 2

See also