Evaluate $\sum_{n=1}^\infty \arctan\left(\frac{1}{8n^2} \right)$ 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 ?.

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