# Sum of an infinite series involving arctan

I came across the following infinite series:

$$\sum_{r=1}^\infty \arctan(\frac{1}{2r^2})$$

I multiplied and divided the argument of arctan by $$2$$ and rewrote the expression as

$$\sum_{r=1}^\infty \arctan(\frac{(2r+1)-(2r-1)}{1 +(2r+1)(2r-1)})$$

Then using the identity $$\arctan(\frac{x - y}{1+xy}) = \arctan(x)-\arctan(y)$$ , got this.

$$\sum_{r=1}^\infty [\arctan(2r+1) - \arctan (2r-1)]$$

Opening this up , we get

$$\arctan(3)-\arctan(1) + \arctan(5) - \arctan(3) ........$$

All the terms except $$-\arctan(1)$$ will get canceled. So the required sum should be $$-\frac{\pi}{4}$$. But in our original series the argument of arctan is between $$\frac{1}{2}$$ and $$\infty$$ which implies the arctan is positive. So the sum should be positive. Did i do anything wrong? I also can not find any other way of summing this.

Any help is appreciated.

• $\lim_{n\rightarrow\infty}\arctan(n)\neq 0$... Unconvergent telescope? Nov 3, 2022 at 17:46
• @BobDobbs the limit is not zero but it's still finite, and that's what matters with such telescopes Nov 3, 2022 at 17:59
• Maybe there is another way... Nov 3, 2022 at 18:03

You are wrong about the terms that get cancelled. You are missing the $$\arctan(\infty)$$ term. It would be more obvious if you write your sum as$$\sum_{r=1}^\infty [-\arctan(2r-1)+\arctan(2r+1)]$$ This is $$\pi/2$$. So the sum will be equal to $$\arctan(\infty)-\arctan(1)=\frac\pi2-\frac\pi 4=\frac\pi 4$$
• When $r$ is very large you keep adding and subtracting $\pi/2$. The subtraction will cancel the $\pi/2$ from the previous term, but you add a new $\pi/2$. In fact, just write some 4 terms:$$-\arctan(1)+\arctan(3)-\arctan(3)+\arctan(5)-\arctan(5)+\arctan(7)-\arctan(7)+\arctan(9)=-\arctan(1)+\arctan(9)$$ Nov 3, 2022 at 17:36
• But that new $\pi/2$ can be canceled by next and so on Nov 3, 2022 at 17:38
• Value of $\pi/4$ confirmed with numpy.
• Think of an infinite sum as the limit of the sequence of its partial sums. At step $n$ your partial sum will be equal to $\arctan(2n+1) - \arctan(1) = \arctan(2n+1) - \pi/4$, and so taking the limit on that sequence gives $\pi/2 - \pi/4 = \pi/4$. Nov 3, 2022 at 17:56