# How to sum the following series?

I am struggling to sum the following series: $$\sum_{n=1}^{\infty}\arctan\left(\frac{2}{n^2}\right).$$ I am not able to start the problem. I guess, any intial hint would be helpful for me. Thanks in advance.

• – QED
Commented Jan 12, 2018 at 7:07
• Hint:$\arctan^{-1}x+\arctan^{-1}y=\arctan^{-1 }(\frac{x+y}{1-xy})$ Commented Jan 12, 2018 at 7:26

Note that $$\arctan\left(\frac{2}{n^2}\right)=\arctan\left(\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\right)=\arctan(n+1)-\arctan(n-1).$$ Hence, for $N\geq 2$, $$\sum_{n=1}^N\arctan\left(\frac{2}{n^2}\right)=\arctan(N+1)+\arctan(N)-\arctan(1)-\arctan(0).$$ Can you take it from here?
• Sir, I have a doubt. If we expand the above series, then it will look something like $tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)....+tan^{-1}(n+1)-tan^{-1}(n-1)$ which will give us $-\frac{\pi}{4}+\frac{π}{2}$ which is equal to $\frac{\pi}{4}$. But textbook is giving the answer $\frac{3\pi}{4}$. Am i doing something wrong while expanding the series. Commented Feb 9, 2018 at 4:40