Find The Sum Involving The Inverse Tangent Function

We are given to evaluate:

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


My solution:

Using the identity:

[math]\tan^{-1}(x)=\cot^{-1}\left(\frac{1}{x} \right)[/math]

we may write:

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

Now, using the fact that:

[math]k^2+k+1=\frac{k(k+1)+1}{(k+1)-k}[/math]

and the identity:

[math]\cot(\alpha-\beta)=\frac{\cot(\alpha)\cot(\beta)+1}{\cot(\beta)-\cot(\alpha)}[/math]

We may now write:

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

This is a telescoping series, hence:

[math]S_n=\cot^{-1}(0)-\cot^{-1}(n+1)=\frac{\pi}{2}-\cot^{-1}(n+1)[/math]

Using the identity:

[math]\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}[/math]

We may also write:

[math]S_n=\tan^{-1}(n+1)[/math]

And so we have found:

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