# How to calculate the series $\sum\limits_{n=1}^{\infty} \arctan(\frac{2}{n^{2}})$?

I encountered the series $$\sum_{n=1}^{\infty} \arctan\frac{2}{n^{2}}.$$

I know it converges (by ratio test), but if I need to calculate its limit explicitly, how do I do that? Any hint would be helpful..

Note that $\arctan(u)-\arctan(v)=\arctan\left(\frac {u-v}{1+uv}\right)$. Taking $u=n+1$ and $v=n-1$ shows that $$\arctan\left(\frac {2}{n^2}\right)=\arctan(n+1)-\arctan(n-1)$$
Thus we see that the series telescopes and $$\sum_{n=1}^{\infty}\arctan\left(\frac {2}{n^2}\right)=2\arctan(\infty)-\arctan(0)-\arctan(1)=\pi -0-\frac {\pi}4=\frac {3\pi}4$$
• @ep pi please explain how you get $\displaystyle \arg \prod^{\infty}_{n=1}\bigg(1-\frac{2i}{n^2}\bigg)$ and also explain second last line., Thanks – DXT Feb 27 '17 at 9:56
• @Durgesh Tiwari The inspiration for this infinite product comes from $arg\;\left(1-\frac{2i}{n^2}\right)=-\arctan(2/n^2)$ and $arg\prod_n r_n e^{\theta_n}=\sum_n\theta_n$. What I did in the second to last line occurs due to the branch I had chosen to work on. – dxdydz Mar 3 '17 at 0:20
• I have made an error in my previous comment, it should be $r_n e^{i\theta_n}%$ under the product, not $r_n e^{\theta_n}$. – dxdydz Mar 3 '17 at 17:10