+179 Compute
$\sum^{2020}_{n=1} [\frac{1}{n^2} - \frac{1}{(n+1)^2}] $
+179 Never mind this equation seems exceedingly obvious unless there's some trick I'm missing.
$\begin{align*} \sum^{2020}_{n=1} [\frac{1}{n^2} - \frac{1}{(n+1)^2}] = \frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{9} + \frac{1}{9} ... -\frac{1}{2020^2} + \frac{1}{2020^2} - \frac{1}{2021^2} \end{align*} $
Thus, terms cancel and..
$$\begin{align*} 1 - \frac{1}{2021^2} \end{align*} $$
