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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 $C_r$?

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    $\begingroup$ Do you have any reason to think there should be? $\endgroup$
    – Greg Martin
    Commented May 2, 2012 at 4:05
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    $\begingroup$ @Greg Martin 1) There do exist reasonably elementary evaluations of $\zeta(2)$ (no Fourier analysis, no contour integration, etc.) so perhaps it's possible to unwind one of them. 2) Any combinatorialist presented with a combinatorial identity naturally wonders whether a bijective proof exists - I don't know of any serious candidates for natural combinatorial facts with no natural proofs. $\endgroup$
    – David Feldman
    Commented May 2, 2012 at 5:46
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    $\begingroup$ Interesting question. Can you spell out the calculus a bit? I see various starts but not a clear path. $\endgroup$
    – Aaron Meyerowitz
    Commented May 2, 2012 at 7:11
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    $\begingroup$ Interesting question, does this generalize to $\zeta(2k)$? $\endgroup$
    – user22202
    Commented May 2, 2012 at 7:16
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    $\begingroup$ Roughly (and not rigorously), each pair $A,B$ contains a first point reading clockwise, say $A$. The condition $|\,||A||−||B||\,|<1$ on $B$ in ${\Bbb R^2}$ locates $B$ in a half-annulus of area about $2\pi||A||$. This half-annulus contains a number of lattice points about equal to it's area, and of those, about $1/\zeta(2)$ satisfy $A⇆B$. Now do a double integral for the possible locations of $A$, first by norm, then by angle. The proportion of points where $A⇆B$ may not behave so will in a thin region, but some integration by parts makes everything legal. $\endgroup$
    – David Feldman
    Commented May 2, 2012 at 22:57

1 Answer 1

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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. $\quad\blacksquare$

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