## Thursday, April 23, 2015

### Find The Sum Involving The Inverse Tangent Function

We are given to evaluate:

$S_n=\sum_{k=0}^n\left[\tan^{-1}\left(\frac{1}{k^2+k+1} \right) \right]$

My solution:

Using the identity:

$\tan^{-1}(x)=\cot^{-1}\left(\frac{1}{x} \right)$

we may write:

$S_n=\sum_{k=0}^n\left[\cot^{-1}\left(k^2+k+1 \right) \right]$

Now, using the fact that:

$k^2+k+1=\frac{k(k+1)+1}{(k+1)-k}$

and the identity:

$\cot(\alpha-\beta)=\frac{\cot(\alpha)\cot(\beta)+1}{\cot(\beta)-\cot(\alpha)}$

We may now write:

$S_n=\sum_{k=0}^n\left[\cot^{-1}(k)-\cot^{-1}(k+1) \right]$

This is a telescoping series, hence:

$S_n=\cot^{-1}(0)-\cot^{-1}(n+1)=\frac{\pi}{2}-\cot^{-1}(n+1)$

Using the identity:

$\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}$

We may also write:

$S_n=\tan^{-1}(n+1)$

And so we have found:

$\sum_{k=0}^n\left[\tan^{-1}\left(\frac{1}{k^2+k+1} \right) \right]=\tan^{-1}(n+1)$