The Basel Problem: How Euler Found π²/6 in the Sum of 1/n²

The Basel problem asks for the exact value of 1 + 1/4 + 1/9 + 1/16 + …, the sum of the reciprocals of the square numbers, and the answer is π²/6, about 1.6449. In 1735 the 28-year-old Leonhard Euler found it by treating sin(x)/x as an infinite polynomial and building it from its roots. π appears because the sine wave’s evenly spaced zeros put squares into its factors.

Add up the reciprocals of the square numbers: one, a quarter, a ninth, a sixteenth, forever. There is no circle anywhere in that question. So why is the answer π² over 6?

Let’s just add. The partial sums climb quickly at first: 1, 1.25, roughly 1.36. Then they crawl. After a hundred terms the sum is roughly 1.635. After ten thousand it is about 1.6448. The series converges, but to what? Those digits look stubbornly unfamiliar, and brute force is hopeless: accuracy to about ten decimal places takes roughly ten billion terms. Adding can estimate the answer. It cannot explain it.

Ninety years of guessing

In 1644 the Italian priest and mathematician Pietro Mengoli posed the question. He could sum other series exactly, but this one resisted. Compare an easier cousin, the sum of 1/(n(n + 1)). Each term splits into 1/n − 1/(n + 1), and when you add them neighboring fractions cancel, so the sum telescopes to exactly 1. The squares offer no such escape.

Forty-five years later the mystery was in Basel, with Jacob Bernoulli. He proved the sum was less than 2, which was useful but not the exact answer. In 1689 he published an appeal: find this sum and send it to us, and we shall be much obliged. The problem acquired his city’s name. Leibniz tried. De Moivre tried. Stirling calculated about a dozen decimal places. But decimals were not an exact evaluation, and for ninety years whole numbers and squares guarded a secret involving circles.

Euler’s audacious move

In 1735 a 28-year-old from Basel, working in Saint Petersburg, announced the answer. Leonhard Euler said the sum was π²/6. Numerically, π² is about 9.8696, and dividing by 6 gives 1.644934, exactly where those partial sums were heading. His reasoning was brilliant, but its boldest step needed justification.

His method begins somewhere unexpected: a polynomial. Suppose it equals 1 at 0 and has roots a and b. Its factors are 1 minus x over each root. Multiply out, and the coefficient of x is minus the sum of the reciprocals of those roots. Roots downstairs, coefficients upstairs. That connection is the whole idea.

Now comes the leap. Take sin(x)/x, with its limiting value 1 at zero. Its Taylor series begins 1 − x²/6 + x⁴/120 and continues forever: something like an infinite polynomial. Its zeros are ±π, ±2π, ±3π, onward.

Euler treated it like a finite polynomial and built an infinite product from its roots. Pair each positive root with its negative twin, and the factors become 1 minus x² over the corresponding root squared. To make an x² term, choose x² from one factor and 1 from every other factor. Its coefficient is therefore minus the sum of 1/π², 1/(4π²), and so on.

But the Taylor series already gave that coefficient: minus one sixth. Equate the two, multiply by −π², and the reciprocal squares sum to π²/6. Ninety years answered in a page.

Was it legal?

The unease was justified. A finite polynomial is fixed by its roots, their multiplicities and a normalization. Infinite series are trickier. Multiply sin(x)/x by e^x²: its zeros stay put and its value at zero remains 1, yet the function changes. Roots alone cannot identify it. Euler’s product is correct, but proving it requires more than listing zeros.

Roughly a century and a half later Karl Weierstrass supplied a general factorization theorem for entire functions. It builds products from zeros, but allows an extra exponential factor, and growth restrictions and symmetry help to determine that missing factor here. Sin(x)/x grows slowly enough in the complex plane for this later theory to justify Euler’s leap.

Euler himself wasn’t idle. By 1741 he had several independent proofs, including arguments using integrals. The answer did not have to wait for the later machinery.

Lighthouses on a circle

A second route is geometric. For ideal point sources, brightness falls as one over distance squared. Put equally strong lighthouses along a straight shoreline, at distances 1, 2, 3 and onward from an observer, and choose units so that their total brightness is our sum.

Now start instead with one lighthouse on a circle, diametrically opposite the observer. An inverse Pythagorean identity lets us replace that light with two carefully positioned lights on a circle twice as wide, and the total brightness stays unchanged. Repeat: double the circle, double the lights, preserve the brightness. Choose the original diameter as 2/π, so that its circumference is 2. As the circles grow, their shoreline flattens, and the lights approach every odd integer distance on both sides of the observer.

The original light’s brightness was π²/4, and that equals twice the sum over the odd squares. So the odd-square sum is π²/8. The even terms contribute a quarter of the whole sum, so the odd terms contribute three quarters. Three quarters of the total equals π²/8, and the total is π²/6. An infinite polynomial and a ring of lights agree.

The door to zeta

Euler didn’t stop at squares. Replace the exponent 2 with s: the sum of 1/n^s. For real s greater than 1 this converges, and we now call it zeta of s. Our answer is ζ(2). Euler also found ζ(4) = π⁴/90, and ζ(6) = π⁶/945. Every positive even integer gives a rational multiple of the corresponding power of π.

And ζ(3)? Nearly three centuries later no comparable closed form is known. In 1978 Roger Apéry astonished mathematicians by proving it irrational. Meanwhile Riemann extended zeta into the complex plane, apart from a pole at 1, and its zeros encode the distribution of the primes.

Mengoli’s question, Bernoulli’s plea, Euler’s leap, Weierstrass’s framework, Riemann’s primes: one innocent sum, and centuries of consequences. Look hard enough at an innocent question, and something round is usually hiding inside it.