Arctangent sum $\sum_{n=1}^{\infty}\arctan\frac{1}{8n^{2}}$ 
Prove the following by considering a telescoping series
  $$\sum_{n=1}^{\infty}\arctan\frac{1}{8n^{2}} = \frac{\pi}{4} - \arctan \left(\tanh\frac{\pi}{4}\right)$$

I recognise this question has been asked here Evaluate $\sum_{n=1}^\infty \arctan\left(\frac{1}{8n^2} \right)$ however, I would specifically like an answer using telescoping given that this question is in the telescoping chapter of Jack D'aurizio's Superior Mathematics from an Elementary Point of View. 
I have tried writing the sum as 
$$\sum_{n=1}^{\infty}\arctan\left(\frac{1}{4n-1}\right) - \arctan\left(\frac{1}{4n+1}\right)$$
but this doesn't telescope and I have no idea how $e$, let alone $\tanh$ is going to appear. The RHS can also be rewritten as $\arctan(e^{-\pi/2})$, but this has not been useful to me.
Could someone provide some hints or an outline of a solution using telescoping?
 A: I won't spoil the solution through creative telescoping, just point that such kind of series can be brute-forced through Weierstrass products. Indeed 
$$\sum_{n\geq 1}\arctan\frac{1}{8n^2} = \text{Im}\sum_{n\geq 1}\log\left(1+\frac{i}{8n^2}\right) = \text{Im}\log\prod_{n\geq 1}\left(1+\frac{i}{8n^2}\right), $$
hence the knowledge of the explicit value of $\prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right)$ is enough to solve the given problem, and
$$ \prod_{n\geq 1}\left(1+\frac{z^2}{n^2}\right) = \frac{\sinh(\pi z)}{\pi z} $$
can be derived from the Weierstrass product of the sine function via the substitution $x\mapsto iz$.

An alternative approach. We have
$$ \arctan\frac{1}{8n^2}=\int_{0}^{\frac{1}{8n^2}}\frac{dx}{1+x^2}=\int_{8n^2}^{+\infty}\frac{dx}{1+x^2}=2\int_{2\sqrt{2}n}^{+\infty}\frac{x\,dx}{1+x^4}=16\int_{n}^{+\infty}\frac{x\,dx}{1+64 x^4} $$
so
$$\begin{eqnarray*} \sum_{n\geq 1}\arctan\frac{1}{8n^2}&=&16\int_{0}^{+\infty}\frac{x\lfloor x\rfloor}{1+64x^4}\,dx=\frac{\pi}{4}-16\int_{0}^{+\infty}\frac{x\{x\}}{1+64x^4}\,dx\\&=&\int_{0}^{+\infty}\frac{16 x}{1+64x^4}\left(\tfrac{1}{2}-\{x\}\right)\,dx\end{eqnarray*}$$
and we may regard $\frac{1}{2}-\{x\}$ as a Fourier sine series:
$$\left(\tfrac{1}{2}-\{x\}\right)\stackrel{\text{a.e.}}{=}\sum_{n\geq 1}\frac{\sin(2\pi n x)}{\pi n}$$
then invoke
$$ \int_{0}^{+\infty}\frac{16x}{1+64x^4}\cdot\frac{\sin(2\pi nx)}{\pi n}\,dx = \frac{\sin\left(\frac{\pi n}{2}\right)}{n e^{\pi n/2}}$$
which follows from the residue theorem or the (inverse) Laplace transform. These manipulations allow to convert the original series into
$$ \sum_{n\geq 1}\frac{\sin\left(\frac{\pi n}{2}\right)}{n e^{\pi n/2}}=\sum_{m\geq 0}\frac{(-1)^m}{(2m+1)e^{\pi(2m+1)/2}} $$
which can be readily computed from $\sum_{m\geq 0}\frac{(-1)^m}{2m+1}z^m=\arctan(z)$, leading to
$$ \sum_{n\geq 1}\arctan\frac{1}{8n^2} = \arctan(e^{-\pi/2}).$$
