# sum of an arctan series using mathematical induction

How to solve this problem using mathematical induction: $$\arctan (1) + \arctan \Big(\frac13\Big) + ... + \arctan \bigg(\frac{1}{n^2+n+1}\bigg)=\arctan (n+1)$$

Hint: $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$
In your case, for the base case $n=1$, $$\tan\left(\arctan 1+\arctan\frac{1}{3}\right)=\frac{\tan(\arctan 1)+\tan(\arctan\frac{1}{3})}{1-\tan(\arctan1)\tan(\arctan\frac{1}{3})}=\frac{1+\frac{1}{3}}{1-\frac{1}{3}}=2$$So $\arctan 1+\arctan\frac{1}{3}=\arctan 2$.