2006 SMT/Algebra Problems/Problem 8

Problem

Evaluate:

$\sum_{x=2}^{10}\frac{2}{x(x^2-1)}.$

Solution

Since the denominator of the fraction is factorable, we try partial fraction decomposition. Let $\frac{2}{(x-1)(x)(x+1)}=\frac{A}{x-1}+\frac{B}{x}+\frac{C}{x+1}$ . Therefore, $2=A(x)(x+1)+B(x-1)(x+1)+C(x-1)(x)$ . Plugging in $x=-1, 0,$ and $1$ , we find that $(A,B,C)=(1,-2,1)$ .

Thus, $\sum_{x=2}^{10}\frac{2}{x(x^2-1)}=\sum_{x=2}^{10}\left(\frac{1}{x-1}-\frac{2}{x}+\frac{1}{x+1}\right)$ . Expanding, we find that this is equal to

$\frac{1}{1}-\frac{2}{2}+\frac{1}{3}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\cdots+\frac{1}{8}-\frac{2}{9}+\frac{1}{10}+\frac{1}{9}-\frac{2}{10}+\frac{1}{11}$ .

We now see that everything cancels (two $\frac{1}{n}$ s cancel with one $-\frac{2}{n}$ ) except for a $\frac{1}{1}, -\frac{2}{2},$ and $\frac{1}{2}$ in the beginning, and a $\frac{1}{10}, -\frac{2}{10}$ , and $\frac{1}{11}$ in the end.

Therefore, our final answer is $\frac{1}{1}-\frac{2}{2}+\frac{1}{2}+\frac{1}{10}-\frac{2}{10}+\frac{1}{11}=\boxed{\frac{27}{55}}$ .

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2006 SMT/Algebra Problems/Problem 8

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