AoPS Wiki talk:Problem of the Day/June 25, 2011

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AoPSWiki:Problem of the Day/June 24, 2011

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The sum in general form is $\sum_{i=1}^{\infty}\frac{1}{(3n-1)(3n+2)}$ We wish to write $\frac{1}{(3n-1)(3n+2)}$ as $\frac{A}{3n-1} + \frac{B}{3n+2}$ We that by setting them equal that $n(3A + 3B) = 0$ and $2A-B=1$ . We find that $A=\frac{1}{3}$ and $B=\frac{-1}{3}$ We now note that $\frac{\frac{1}{3}}{3n-1} - \frac{\frac{1}{3}}{3n+2} = \frac{1}{(3n-1)(3n+2)}$ We now rewrite the sum as $\sum_{i=1}^{\infty}\frac{\frac{1}{3}}{3n-1} - \frac{\frac{1}{3}}{3n+2}$ Now we note that $3(n+1)-1=3n+3-1=3n+2$ . As a result, this is a telescoping sum. Hence, the total sum is the first term or $\boxed{\frac{1}{6}}$

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AoPS Wiki talk:Problem of the Day/June 25, 2011

Problem

Solutions