I have to find $$\sum_{n=1}^\infty \operatorname{arccot}(n^2-n+1)$$ and trying to compute sum of this series. I'm trying to take the derivative of member of the series and compute another sum, but this is not very successful.
Is it right way to try take the derivative or it is useless in this situation? I can't see how to simplify this monster.
Use $ \cot^{-1}(x) = \tan^{-1}(1/x)$ and \begin{eqnarray*} \tan^{-1}(A)-\tan^{-1}(B) =\tan^{-1} \left( \frac{A-B}{1+AB} \right) \end{eqnarray*} We have \begin{eqnarray*} \sum_{n=1}^\infty \cot^{-1}(n^2-n+1) &=& \sum_{n=1}^\infty \tan^{-1} \frac{1}{n^2-n+1} \\ &=& \sum_{n=1}^\infty \tan^{-1} \frac{n-(n-1)}{1+n(n-1)} \\ &=& \sum_{n=1}^\infty \left(\tan^{-1} \left(\frac{1}{n-1}\right) +\tan^{-1} \left(\frac{1}{n} \right) \right)= \color{red}{\frac{\pi}{2}}. \\ \end{eqnarray*}
