how to calculate $\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}$

Question

Question: how to calculate $$\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}$$

My attempt

Let
$\arctan \theta =\frac{i}{2}\ln \left( \frac{i+\theta }{i-\theta } \right)$

$$S=\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\arctan \frac{1}{1+k^{2}}}=\underset{n\to +\infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\frac{i}{2}\ln \left( \frac{i+\frac{1}{1+k^{2}}}{i-\frac{1}{1+k^{2}}} \right)}$$

$$=\frac{i}{2}\underset{n\to +\infty }{\mathop{\lim }}\,\ln \prod\limits_{k=1}^{n}{\left( \frac{i+\frac{1}{1+k^{2}}}{i-\frac{1}{1+k^{2}}} \right)}=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{i+\frac{1}{1+k^{2}}}{i-\frac{1}{1+k^{2}}} \right)} \right)$$

$$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{i\left( 1+k^{2} \right)+1}{i\left( 1+k^{2} \right)-1} \right)} \right)=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{1+k^{2}-i}{1+k^{2}+i} \right)} \right)$$

$$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=1}^{n}{\left( \frac{k^{2}+1-i}{k^{2}+1+i} \right)} \right)=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=0}^{n}{\left( \frac{\left( k+1 \right)^{2}+1-i}{\left( k+1 \right)^{2}+1+i} \right)} \right)$$

Note that $$\left( k+1 \right)^{2}+1\pm i=\left( k+1+i\sqrt{1\pm i} \right)\left( k+1-i\sqrt{1\pm i} \right)$$ $$=\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\frac{\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1-i} \right)}}{\prod\limits_{k=0}^{n}{\left( k+1+i\sqrt{1+i} \right)}}\cdot \frac{\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1-i} \right)}}{\prod\limits_{k=0}^{n}{\left( k+1-i\sqrt{1+i} \right)}} \right)$$

Answer 1

It's good you did all that ground work. Let's rewrite

$$\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=0}^{n}{\left( \frac{\left( k+1 \right)^{2}+1-i}{\left( k+1 \right)^{2}+1+i} \right)} \right) = \frac{i}{2}\ln\; \left(\frac{\prod_{k=1}^\infty 1 - \frac{i-1}{k^2}}{\prod_{k=1}^\infty 1 - \frac{-i-1}{k^2}}\right) \qquad (1)$$ where each of the products on the right is convergent. To deal with this, use the product formula for the sine function:

$$\frac{\sin \pi x}{\pi x} = \prod_{k=1}^\infty \left(1 - \frac{x^2}{k^2}\right).$$ We'll apply this in a moment, but first let's observe that right side of (1) is of the form $\frac{i}{2} \ln \frac{z}{(\overline{z})} = -\theta$, where $z = re^{i\theta}$ (using throughout the convention $-\pi < \theta \leq \pi$). So it is only a matter of identifying the $\theta$-coordinate of the product in the numerator appearing in (1).

Applying the product formula to the numerator in (1), setting $x^2 = i - 1 = \sqrt{2}\exp(\frac{3\pi i}{4})$ so that $x = 2^{1/4}\exp(\frac{3\pi i}{8})$ (up to a sign), the numerator is thus

$$\frac{\sin \pm \pi 2^{1/4} \exp(\frac{3\pi i}{8})}{\pm \pi 2^{1/4} \exp(\frac{3\pi i}{8})}$$ and of course the sign won't matter since $\frac{\sin \pi x}{\pi x}$ is an even function. The $\pi 2^{1/4}$ in the denominator can be discarded since we're only interested in the $\theta$-coordinate. Incorporating the angle for that denominator, our answer will thus be $\frac{3\pi}{8} - \theta(x, y)$ where

$$\theta(x, y) = \arccos \frac{x}{(x^2 + y^2)^{1/2}},$$ here putting

$$x + iy = \sin\left(\pi 2^{1/4} \left(\cos \frac{3\pi}{8} + i\sin \frac{3\pi}{8}\right)\right)$$ which, judging by your work above, it looks like you'll be quite capable of handling in terms of hyperbolic functions and whatnot.

Answer 2

Note that by the Weirestrass factorsation of $\sin(z)$, $$\sinh(\pi z) = \pi z\prod_{k=0}^{\infty} \left(1+\frac{z^2}{(k+1)^2}\right)$$

Thus the expression you left with, becomes $$\frac{i}2 \ln\left(\frac{\sqrt{1+i}}{\sqrt{1-i}}\cdot\frac{\sinh\left(\pi\sqrt{1-i}\right)}{\sinh\left(\pi\sqrt{1+i}\right)}\right)= \frac{-\pi}{8}+\frac{i}2\ln(\sinh(\pi\sqrt{1-i})\text{csch}(\pi\sqrt{1+i}))$$

I'm not going to try to simplify the part with the $\ln$, but you could surely give it a try to see if it simplifies further. It seems unlikely though given a pesky $\sqrt[4]{2}$ factor lingering in with $e^{i\pi/8}$ because of $\sqrt{1\pm i}$.

Numerically the sum evaluates to $\approx 1.037$.