(Approximately) bijective proof of $\zeta(2)=\pi^2/6$? Given $A,B\in {\Bbb Z}^2$, write $A \leftrightarrows B$ if the
interior of the line segment $AB$ misses
${\Bbb Z}^2$.
For $r>0$, define
$S_r:=\{ \{A, B\} \mid  A,B\in {\Bbb Z}^2,\|A\|<r,\|B\|<r, \left| \|A\|-\|B\| \right|<1 \text{ and } A \leftrightarrows B \}\ .$
A little calculus gives the equivalence of $\zeta(2)=\pi^2/6$ and
$$\lim_{r\to\infty} \frac{|S_r|}{(2r)^3} = 1\ .$$
Of course $(2r)^3$ counts lattice points in a cube $C_r:=[-r,r)^3$.
Question: Does there exist an approximately bijective proof of this
limit (or some variant), one that matches most of $S_r$ with most of $C_r$?
 A: A proof of $\displaystyle \sum_{n=1}^\infty \frac 1{n^2} = \frac{\pi^2}6$ is described in a youtube video that I may find and post later today, after it appeared in a paper that I may cite here later today, so I'll let this present posting serve as a reminder to get back to this when I have more time. (As I thought might happen, someone beat me to the URL for the video. See the comment below.) (Another postscript: The paper is by Johan Wästlund. Here is the preprint.)
This involves a one-to-two correspondence (so just construe "bi-" in "bijective" as referring to that "two").
It goes like this. First observe that the proposition to be proved is easily seen to be equivalent to $\displaystyle \sum_{\text{odd } n\,\in\,\mathbb Z} \frac 1{n^2} = \frac{\pi^2}4 $.
Next, approximate this last sum by the sum of squares of the reciprocals of the distances from $0$ of certain points on the circle of circumference $2^n$ that touches the line $y=0$ in the plane $(x,y)\in\mathbb R^2$ at the point $(0,0)$. Those specified points are the those whose distance measured along the circle from $(0,0)$ is an odd integer.
The central lemma is that if $n$ increases by $1$, then that sum does not change. This reduces the problem to the the case $n=1$, and in that case the sum is $\pi^2/4$.
The proof of the lemma is from secondary-school geometry: Through each of the specified points draw the line through the very top of the circle. That line intersects the next circle, the case $n+1$, at two points. Show that those are two of the points at odd-integer arc lengths from $(0,0)$ and that the sum of the squares of the reciprocals of the distances from those two points to $(0,0)$ is equal to the reciprocal of the square of the distance to $(0,0)$ from the point you started with. $\quad\blacksquare$
