HMMT February

Each term takes the form $$\frac{1}{n^{2}+(n-2)}=\frac{1}{(n+2) \cdot(n-1)}$$ Using the method of partial fractions, we can write (for some constants $A, B$ ) $$\begin{gathered} \frac{1}{(n+2) \cdot(n-1)}=\frac{A}{(n+2)}+\frac{B}{(n-1)} \\ \Rightarrow 1=A \cdot(n-1)+B \cdot(n+2) \end{gathered}$$ Setting $n=1$ we get $B=\frac{1}{3}$, and similarly with $n=-2$ we get $A=-\frac{1}{3}$. Hence the sum becomes $$\frac{1}{3} \cdot\left[\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{3}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\cdots\right]$$ Thus, it telescopes, and the only terms that do not cancel produce a sum of $\frac{1}{3} \cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{13}{36}$.