Prove that $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$ Prove that 
$$ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} =  \tan^{-1}\frac{n}{n+2}$$
I have been trying to solve it step by like $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7}=\tan^{-1}\frac{1}{2}$ and so on but cannot observe any pattern. Could someone suggest something?

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