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?

  • 1
    $\begingroup$ maybe using $\arctan(1/2k^2) = \int_0^{1/2k^2} \frac{dx}{1+x^2} = \int_0^{1} \frac{dx}{2 k^2(1+ (x/2k^2)^2)} = \frac{1}{2}\int_0^{1} \frac{dx}{2k^2+ x/2k^2} = \int_0^{1} \frac{dx}{(2k+i x/2k)(2k-i x/2k)} = \frac{1}{4k}\int_0^{1} \left(\frac{1}{2k+ix/2k} + \frac{1}{2k-ix/2k}\right)dx$ and if reduces to computing $\sum_k \frac{1}{8k^2+ix}$ probably using a known formula for $\cot(z)$ $\endgroup$ – reuns Apr 24 '16 at 8:24
  • $\begingroup$ The students do not know integration right now. And certainly complex analysis is not in the syllabus. I like the solution though, thanks! $\endgroup$ – Isomorphism Apr 24 '16 at 15:44

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.


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