Geometric Interpretation of the Basel Problem? Does the Pi in the solution to the Basel problem have any geometric significance?  Every time I see Pi, I have to think of a circle.  I would love to see a nice intuitive picture connecting the Basel problem with geometrical figures.  Anyone here been lucky enough to stumble upon such a thing?

The Montreal interpretation of the Basel problem
Suppose you randomly pick a street in the left window, one which is parallel to that window (red streets). Similarly, you pick a blue street in the right window: 

Then, from this perspective the cross section between the two streets has 50% chance of being visible in the right window (the blue street is farther away than the red street).
Hence:
$$ \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} $$
Proof
Okay, I assumed that in each window the elevation angle of the picked street was uniformly distributed between $0^\circ$ and $90^\circ$. This means that the distances are Cauchy distributed: 
$$P_{\color{blue}{X}}(\color{blue}{x}) = \frac{2}{\pi (1+\color{blue}{x}^2)}$$
$$P_{\color{red}{Y}}(\color{red}{y}) = \frac{2}{\pi (1+\color{red}{y}^2)}$$
NB: The unit distance of $\color{blue}{x}$ and $\color{red}{y}$ is the viewer’s height above the ground.
The statement that the cross section is in 50% of the cases visible in the right window, is equivalent to saying that the ratio $\frac{\color{red}{y}}{\color{blue}{x}}$ has 50% chance being smaller than 1. Note that this ratio is the tangens of the azimuth where the cross section is located. 
A ratio of two similar Cauchy distributions, is distributed as follows: 
$$P_T(t) = \frac{4}{\pi^2}\frac{\log\left(\frac{1}{t}\right)}{(1 - t^2)}$$
Using $P_T(t<1)=0.5$ , we get: 
$$ \begin{align}\frac12 &= \frac{4}{\pi^2}\int_0^1 \frac{\log\left(\frac{1}{t}\right)}{1 - t^2}dt \\ \frac{\pi^2}{8} &= \sum_{k=0}^\infty \int_0^1 \log\left(\frac{1}{t}\right) t^{2k}dt  \\ \frac{\pi^2}{8} &= \sum_{k=0}^\infty  \left.\frac{t^{2k+1}}{2k+1}\left(\frac{1}{2k+1} - \log(t)\right)\right|_0^1  \\ \frac{\pi^2}{8} &= \sum_{k=0}^\infty \frac1{(2k+1)^2} \\ \frac{\pi^2}{8} &= \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \cdots \end{align}$$
The complete Basel sum
$$\displaystyle \begin{align} \sum_{k=1}^\infty \frac{1}{k^2} &= \frac{4}{3}\left(\sum_{k=1}^\infty \frac{1}{k^2} - \frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^2} \right) \\ &= \frac{4}{3}\left(\sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=1}^\infty \frac{1}{(2k)^2} \right)  \\ &= \frac{4}{3}\left(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \cdots\right) \\ \sum_{k=1}^\infty \frac{1}{k^2} &=\frac{\pi^2}{6} \\ & \qquad \qquad \blacksquare \end{align}$$
Based on a proof by Luigi Pace, as appeared in: ‘The American Mathematical Monthly’, August-September 2011, pp. 641-643
