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2018 CHMMC Tiebreaker 3 - sum ( 1/(n^2 + 3n) - 1/(n^2 + 3n + 2) )

Source:

March 2, 2024
algebraSumCHMMC

Problem Statement

Compute n=1(1n2+3n1n2+3n+2)\sum^{\infty}_{n=1} \left( \frac{1}{n^2 + 3n} - \frac{1}{n^2 + 3n + 2}\right)