Hiperbolic geometry with Riemann function

Geometrically, we can think of this as a sum of the terms $cos^{2}(\theta)$ , where $\theta$ is the internal angle in a triangle within the unit circle with hypotenuse $1$ and sides $nx$ and $x$ .

Setting $\theta = arctan(nx/x)$ , it's easy for us to see that $cos(\theta) = x = \sqrt{1-n^2x^2}$ . So, the sum turns into

$\displaystyle\sum_{n=1}^\infty cos^2(arctan(n)) = \displaystyle\sum_{n=1}^\infty 1-n^2x^2 = \displaystyle\sum_{n=1}^\infty 1 - \frac{n^2}{1+n^2}$

Where the last equation was obtained by using the Pythagorean theorem ( $1 = x^2 + n^2x^2 = x^2(1+n^2)$ ) to solve for $x$ in terms of $n$ . The terms of the sum converge to $0$ as $n \to \infty$ . Additionally, we know that it is also bounded by the geometric series $\sum 1/n^2$ so it seems like we're in good shape for the whole thing to converge. In fact, we can simplify things further to

$\displaystyle\sum_{n=1}^\infty \frac{1}{1+n^2}$

We also have the identity

$\displaystyle\sum_{n=-\infty}^\infty \frac{1}{a^2+n^2} = \frac{\pi}{a}coth(\pi a)$

In order to turn this into the form of the original problem, we have to set $a=1$ and cut the sum into half. This can be done by taking out the $n=0$ term, $1$ , and dividing everything by $2$ . We can only do this since the terms are all the same at both negative and positive values of $n$ . Doing this turns the indices of the sum from $1$ to $\infty$ .

$\frac{1}{2} + \displaystyle\sum_{n=1}^\infty \frac{1}{1+n^2} = \frac{\pi}{2}coth(\pi)$

FInally,

$\displaystyle\sum_{n=1}^\infty \frac{1}{1+n^2} = \frac{\pi}{2}coth(\pi) - \frac{1}{2} = 1.07634$