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: $$\sum_{r=1}^n \tan^{-1}\frac{1}{r^2+r+1}$$
$$=\sum_{r=1}^n \tan^{-1}\frac{(r+1)-r}{1+r(r+1)}$$
$$=\sum_{r=1}^n [\tan^{-1} (r+1) -\tan^{-1} r]$$
