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Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 7"
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Therefore, <math>\frac{4n}{(n^2-1)^2}=\frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</math>, so <math>S =\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</math> <math>= \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} - \sum_{n=2}^{\infty} \frac{1}{(n+1)^2}</math>  
 
Therefore, <math>\frac{4n}{(n^2-1)^2}=\frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</math>, so <math>S =\sum_{n=2}^{\infty} \frac{1}{(n-1)^2}-\frac{1}{(n+1)^2}</math> <math>= \sum_{n=2}^{\infty} \frac{1}{(n-1)^2} - \sum_{n=2}^{\infty} \frac{1}{(n+1)^2}</math>  
  
−
Writing out the first few terms, we have
+
Writing out the first few terms and rearranging, we have
−
<math>\left( \frac{1}{1^2}+\frac{1}{2^2}\cancel{+\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right)- \left(\cancel{+\frac{1}{3^2}+\frac{1}{4^2}+\cdots } \right) = \frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}</math>
+
<math>\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)</math>, which telescopes to <math>\frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}</math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 10:57, 23 October 2023

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

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 $\frac{1}{1^2}+\frac{1}{2^2} = \boxed{\frac{5}{4}}$

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

See also

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
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