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.

  • 1
    $\begingroup$ Actually this proof is contained in Hilbert--Cohn-Vossen's "Geometry and imagination". $\endgroup$ Jul 31, 2018 at 6:34

2 Answers 2


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.

  • 5
    $\begingroup$ What a fantastic proof, nonetheless! $\endgroup$ Dec 21, 2009 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.