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: Observe last term of $L.H.S.$ carefully and note that $\tan^{-1}a-\tan^{-1} b=\tan^{-1}\frac{a-b}{1+ab}$
