# Is there a “finitary” solution to the Basel problem?

Gabor Toth's Glimpses of Algebra and Geometry contains the following beautiful proof (perhaps I should say "interpretation") of the formula $\displaystyle \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} \mp ...$, which I don't think I've ever seen before. Given a non-negative integer $r$, let $N(r)$ be the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 \le r^2$, i.e. the number of lattice points in the ball of radius $r$. Then if $r_2(n)$ is the number of ordered pairs $(a, b) \in \mathbb{Z}^2$ such that $a^2 + b^2 = n$, it follows that $N(r^2) = 1 + r_2(1) + ... + r_2(r^2)$.

On the other hand, once one has characterized the primes which are a sums of squares, it's not hard to show that $r_2(n) = 4(d_1(n) - d_3(n))$ where $d_i(n)$ is the number of divisors of $n$ congruent to $i \bmod 4$. So we want to count the number of divisors of numbers less than or equal to $r^2$ congruent to $i \bmod 4$ for $i = 1, 3$ and take the difference. This gives

$\displaystyle \frac{N(r^2) - 1}{4} = \left\lfloor r^2 \right\rfloor - \left\lfloor \frac{r^2}{3} \right\rfloor + \left\lfloor \frac{r^2}{5} \right\rfloor \mp ...$

and now the desired result follows by dividing by $r^2$ and taking the limit.

Question: Does a similar proof exist of the formula $\displaystyle \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ...$?

By "similar" I mean one first establishes a finitary result with a clear number-theoretic or combinatorial meaning and then takes a limit.

• Actually this proof is contained in Hilbert--Cohn-Vossen's "Geometry and imagination". – Ivan Izmestiev Jul 31 '18 at 6:34

## 2 Answers

I think that the 14th and last proof in Robin Chapman's collection is just that. It relies on the formula for the number of representations of an integer as a sum of four squares, which is kind of overkill, but anyway.

• What a fantastic proof, nonetheless! – Harrison Brown Dec 21 '09 at 22:11

A somewhat different perspective to the Basel problem relates $$\zeta(2)$$ to the volume of $$SL_2(\mathbb{R})/SL_2(\mathbb{Z})=\zeta(2)/2$$. They compute this volume via a count of lattice points. One can also compute this via Gauss-Bonnet as a circle bundle over the modular curve $$\mathbb{H}^2/PSL_2(\mathbb{Z})$$ and deduce the Basel identity. There is some subtlety here about how the volume forms are defined in the comparison, but I think that this can be made into a proof.