Please calculate $\sum _{ k=0 }^\infty\left[ \tan^{ -1 }\left( \frac { 1 }{ k^{ 2 }+k+1 } \right) -\ldots \right] $

Question

Not many math problems stump me, but this summation has me stumped. Can someone provide a solution to this summation: $$\sum _{ k=0 }^{ \infty }{ \left[ \tan ^{ -1 }{ \left( \frac { 1 }{ k^{ 2 }+k+1 } \right) } -\tan ^{ -1 }{ \left( \frac { 2 }{ k^2+2k+1 } \right) } \right] } $$

Answer 1

Well we have:

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

So:

$$\alpha - \beta = \tan^{-1}\left(\frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\right)$$

Provided $\alpha - \beta$ is in the range $(-\pi/2,\pi/2)$.

If $\alpha = \tan^{-1}(x)$ and $\beta = \tan^{-1}(y)$, this gives:

$$\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right)$$

Which holds for all real numbers with $xy > -1$. In particular,

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

This allows you to evaluate:

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

I'm not sure about the other term. If it were $\tan^{-1}\left(\frac{2}{k^2 + 2k + 1}\right)$ you could do a similar trick:

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

And so:

$$\sum_{k = 0}^{n} \tan^{-1}\left(\frac{2}{k^2 + 2k + 1}\right) = \tan^{-1}(n + 2) + \tan^{-1}(n + 1) - \tan^{-1}(1) - \tan^{-1}(0) \rightarrow \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{4}$$

As it is, I'm stumped. Wolfram Alpha gives a closed form: http://www.wolframalpha.com/input/?i=sum+from+k+%3D+1+to+infinity+of+tan%5E%28-1%29%281%2Fk%5E2%29.