+0  
 
0
70
1
avatar+181 

Compute

 

$\sum^{2020}_{n=1} [\frac{1}{n^2} - \frac{1}{(n+1)^2}] $ 

 Jul 26, 2021
edited by Guest  Jul 26, 2021
 #1
avatar+181 
0

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..

 

$$\begin{align*} 1 - \frac{1}{2021^2} \end{align*} $$

 Jul 26, 2021

39 Online Users

avatar
avatar