2013 Mock AIME I Problems/Problem 10

Problem: Let $T_n$ denote the $n$th triangular number, i.e. $T_n=1+2+3+\cdots+n$. Let $m$ and $n$ be relatively prime positive integers so that \[\sum_{i=3}^\infty \sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\dfrac{m}{n}.\] Find $m+n$.


Solution: Note \[\sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\frac{\frac{3}{T_i}}{1-\frac{3}{T_i}}=\frac{3}{T_i-3}.\] So we wish to evaluate \[\sum_{i=3}^\infty\frac{3}{T_i-3}=\sum_{i=3}^\infty\frac{6}{i^2+i-6}=\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}.\] It is not difficult to check $\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\frac{1}{i-2}-\frac{1}{i+3}\right)$. Telescoping, we obtain \[\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\sum_{i=3}^\infty \frac{1}{i-2}-\frac{1}{i+3}\right)=\frac{6}{5}(1+1/2+1/3+1/4+1/5)=\frac{137}{50}.\] Hence, $m+n=\boxed{187}$.

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