I found this series in Jack D'Aurizio's Superior Mathematics from an Elementary Point of View on his user page. So I've seen similar series to this, so I figured I tried to make it telescope. I managed to write it using the difference formula for $\arctan\left(x\right)$ so,

\begin{align} \sum_{n = 1}^{\infty} \arctan\left(\frac{1}{8n^{2}}\right) & = \sum_{n = 1}^{\infty} \left[\vphantom{\large A}\arctan\left(4n + 1\right) -\arctan\left(4n - 1\right)\right] \\[1mm] & = \sum_{n = 1}^{\infty}\left(-1\right)^{n}\arctan\left(2n + 1\right) \end{align}

Writing the series that though does not seem to help since none of the terms cancel with each other. What should I do find the answer ?.

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    $\begingroup$ That last sum doesn't converge since the absolute value of the terms tend to $\pi/2$ not $0$. The first two sums do converge though. $\endgroup$ – JimmyK4542 Jan 6 '19 at 18:24
  • $\begingroup$ The last equality is not correct... $\endgroup$ – Fabian Jan 6 '19 at 18:25
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    $\begingroup$ Try this approach for a sutable $x$. $\endgroup$ – A.Γ. Jan 6 '19 at 18:31
  • $\begingroup$ I see you guy's point. I put that there since Desmos seemed to say it's equal to the original sum, but it definitely diverges. $\endgroup$ – Tom Himler Jan 6 '19 at 18:45

We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is $-\text{arctan}(1/8n^2)$.


$$\begin{align*} \sum_{n=1}^\infty\text{arctan}\left(\frac{1}{8n^2}\right) &=-\text{arg}\prod_{n=1}^\infty\left(1-\frac{i}{8n^2}\right) \\ &= -\text{arg}\prod_{n=1}^\infty\left(1-\frac{\left(\sqrt{i/8}\right)^2}{n^2}\right)\\ &=-\text{arg}\frac{\sin\left(\pi\sqrt{i/8}\right)}{\pi\sqrt{i/8}},\qquad\text{by Euler's product for the sine} \\ &=-\text{arg}\left(\frac{(1-i)\sqrt{2}\cosh(\pi/4)}{\pi}+\frac{(1+i)\sqrt{2}\sinh(\pi/4)}{\pi}\right) \\ &=-\text{arctan}\left(\frac{\sinh(\pi/4)-\cosh(\pi/4)}{\sinh(\pi/4)+\cosh(\pi/4)}\right) \\ &= \text{arccot}\left(e^{\pi/2}\right). \end{align*}$$

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