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2000 SMT/Advanced Topics Problems/Problem 3

Revision as of 08:27, 24 July 2020 by Orca25 (talk | contribs) (Created page with "==Problem== Evaluate <math>\sum^{\infty}_{n=1}\frac{1}{n^2+2n}.</math> ==SMT Solution== We know that <math>\frac{1}{n^2+2n}=\frac{1}{n(n+2)}=\frac{\frac{1}{n}-\frac{1}{n+...")
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Problem

Evaluate $\sum^{\infty}_{n=1}\frac{1}{n^2+2n}.$


SMT Solution

We know that $\frac{1}{n^2+2n}=\frac{1}{n(n+2)}=\frac{\frac{1}{n}-\frac{1}{n+2}}{2}.$ So, if we sum this from $1$ to $\infty,$ all terms except for $\frac{\frac{1}{1}}{2}+\frac{\frac{1}{2}}{2}$ will cancel out (a "telescoping" series). Therefore, the sum will be $\mathbf{\frac{3}{4}}.$




Credit

Problem and solution were taken from https://sumo.stanford.edu/old/smt/2000/

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