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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.

Thanks in advance.

Vedant Chhapariya
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1 Answers1

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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 $$\arctan(\infty)-\arctan(1)=\frac\pi2-\frac\pi 4=\frac\pi 4$$

Andrei
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  • But the sum is going continuously how can there be anything left at the end? Cause there is no end. – Vedant Chhapariya Nov 03 '22 at 17:34
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    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:$$-\arctan(1)+\arctan(3)-\arctan(3)+\arctan(5)-\arctan(5)+\arctan(7)-\arctan(7)+\arctan(9)=-\arctan(1)+\arctan(9)$$ – Andrei Nov 03 '22 at 17:36
  • But that new $\pi/2$ can be canceled by next and so on – Vedant Chhapariya Nov 03 '22 at 17:38
  • Value of $\pi/4$ confirmed with numpy. – Doug Nov 03 '22 at 17:39
  • Correct. But at some point the difference is so small, so you keep adding zeros. – Andrei Nov 03 '22 at 17:40
  • Exactly i am adding 0s continuously so how can there be anything left at end? – Vedant Chhapariya Nov 03 '22 at 17:42
  • @Andrei for n terms it makes sense to me. But for infinite terms I just can wrap my head around it – Vedant Chhapariya Nov 03 '22 at 17:47
  • Perhaps you can apply the cancellation argument to a sufficiently large partial sum (only need about 12 terms). Then reason that for every two terms added past this point, the change in the sum will be negligible. – Doug Nov 03 '22 at 17:54
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    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 $\pi/2 - \pi/4 = \pi/4$. – Bruno B Nov 03 '22 at 17:56