2016 UNCO Math Contest II Problems/Problem 7


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


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:


See also

2016 UNCO Math Contest II (ProblemsAnswer KeyResources)
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