The following is a popular question (in competitive exams) in India:
Compute the value of $S=\displaystyle \sum_{k=1}^{\infty} \tan^{-1}\left( \dfrac1{2k^2}\right)$.
I can compute the value by writing $$ \tan^{-1}\left( \dfrac1{2k^2}\right) = \tan^{-1} (2k+1) - \tan^{-1} (2k-1).$$
The telescoping sum gives $S =\dfrac{\pi}4$.
However, this question was listed under the chapter titled Complex Numbers. I wanted to know if there was a "complex number" approach to this problem.
My attempts are the following:
1) I can force a complex number proof (motivated by the telescoping inverse trigonometric sum) by looking at the infinite product where the $k$th term is $$\dfrac{1+i(2k+1)}{1+i(2k-1)}$$ which does not seem natural.
2) Another approach is to use $$\displaystyle \sin z = z\prod_{k=1}^{\infty} \left(1-\dfrac{z^2}{k^2 \pi^2}\right). $$ Substituting $$\displaystyle z = \dfrac{(1-i)\pi}{2},$$ we get $$\dfrac{\sin z}{z}= \prod_{k=1}^{\infty} \left(1+\dfrac{i}{2k^2}\right)=P.$$ Notice that the angle of the complex number $P$ is $S$ (the required infinite sum).
It can be shown that $\dfrac{\sin z}{z} = \left(\dfrac{e^\frac{\pi}{2}+e^{-\frac{\pi}{2}}}{2 \pi}\right)\cdot (1+i).$
Since the real and imaginary parts of $P$ are equal, the angle of the complex number $P$ is $\frac{\pi}{4}$.
Is there a simple complex number solution to this series?
Perhaps the intention is to use the following identity:
$$\arctan z=\frac12i\left(\log(1-iz)-\log(1+iz)\right)\;,\;\;z\in\Bbb C$$
Followed, maybe, by the use of the exponential function.