# 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 $\tan^{-1}a-\tan^{-1} b=\tan^{-1}\frac{a-b}{1+ab}$ – Mathematics Apr 22 '16 at 16:18
• Try mathematical induction ! – Mambo Apr 22 '16 at 16:18
• – lab bhattacharjee Apr 22 '16 at 16:30

Hint: Observe last term of $L.H.S.$ carefully and note that $\tan^{-1}a-\tan^{-1} b=\tan^{-1}\frac{a-b}{1+ab}$
HINT: $$\sum_{r=1}^n \tan^{-1}\frac{1}{r^2+r+1}$$ $$=\sum_{r=1}^n \tan^{-1}\frac{(r+1)-r}{1+r(r+1)}$$ $$=\sum_{r=1}^n [\tan^{-1} (r+1) -\tan^{-1} r]$$
Hint: $$\arctan\frac {1}{1+k+k^2}=\arctan\frac{(k+1)-k}{1+(k+1)k}=\arctan(k+1)-\arctan k$$