The Basel problem, and where the π comes from

Why the sum of 1/n² equals π²/6

A proof of the Basel sum using Fourier series and Parseval’s identity, plus an interactive look at how slowly the partial sums converge.

In 1734 Euler announced that \sum_{n=1}^{\infty} \frac{1}{n^2} \;=\; 1 + \frac14 + \frac19 + \frac1{16} + \cdots \;=\; \frac{\pi^2}{6}. \tag{1}

The problem had been open since 1650 and had beaten the Bernoullis. The surprise is the \pi: nothing on the left side mentions circles. Below is a proof that makes it clear where the circle comes from.

Convergence is the easy part

Comparing with a telescoping sum shows that Equation 1 converges, and to something less than 2: \begin{aligned} \sum_{n=1}^{N} \frac{1}{n^2} &\;<\; 1 + \sum_{n=2}^{N} \frac{1}{n(n-1)} \\ &\;=\; 1 + \sum_{n=2}^{N}\left(\frac{1}{n-1} - \frac1n\right) \;=\; 2 - \frac1N. \end{aligned}

Finding the exact value is much harder. Convergence is also slow: the tail after N terms is about 1/N, so after a hundred terms you are still off by about 0.01. Try it:

viewof N = Inputs.range([1, 2000], {value: 25, step: 1, label: "Terms N"})

partial = { let s = 0; for (let n = 1; n <= N; n++) s += 1 / (n * n); return s; }
target = Math.PI ** 2 / 6
html`<div class="am-stat-row">
  <div><span>Partial sum</span><strong>${partial.toFixed(8)}</strong></div>
  <div><span>π²/6</span><strong>${target.toFixed(8)}</strong></div>
  <div><span>Error × N</span><strong>${((target - partial) * N).toFixed(5)}</strong></div>
</div>`

Plot.plot({
  height: 260,
  marginLeft: 50,
  style: {background: "transparent", fontFamily: "Familjen Grotesk, sans-serif", fontSize: "13px"},
  x: {label: "n", grid: false},
  y: {label: "partial sum", domain: [1, 1.7], grid: true},
  marks: [
    Plot.ruleY([target], {stroke: "currentColor", strokeDasharray: "4 4", strokeOpacity: 0.6}),
    Plot.line(d3.range(1, N + 1).map(n => ({n, s: d3.sum(d3.range(1, n + 1), k => 1 / (k * k))})),
      {x: "n", y: "s", stroke: "#e3a21a", strokeWidth: 2.5})
  ]
})

The last number, error × N, approaches 1. That follows from comparing the tail with an integral: \sum_{n > N} n^{-2} \approx \int_N^\infty x^{-2}\dd x = 1/N.

The proof: Fourier series and Parseval’s identity

The idea is to compute one integral in two different ways. The \pi will come from the length of the interval [-\pi, \pi].

Theorem 1 (Parseval) If f \colon [-\pi, \pi] \to \R is square-integrable with Fourier series \frac{a_0}{2} + \sum_{n \ge 1} \big(a_n \cos nx + b_n \sin nx\big), then \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)^2 \dd x \;=\; \frac{a_0^2}{2} + \sum_{n=1}^{\infty} \left(a_n^2 + b_n^2\right).

This is the Pythagorean theorem in infinitely many dimensions: the functions \tfrac{1}{\sqrt 2}, \cos nx, \sin nx form an orthonormal basis for the inner product \inner{f}{g} = \frac1\pi\int_{-\pi}^{\pi} fg, and the squared length of a vector is the sum of the squares of its coordinates.

Theorem 2 (Euler, 1734) \displaystyle\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{6}.

Proof. Take f(x) = x on [-\pi, \pi]. It’s odd, so a_n = 0 for all n. Integrating by parts, \begin{aligned} b_n \;=\; \frac{1}{\pi}\int_{-\pi}^{\pi} x \sin nx \dd x &\;=\; \frac{2}{\pi}\left[-\frac{x\cos nx}{n}\right]_0^{\pi} + \frac{2}{\pi n}\int_0^\pi \cos nx \dd x \\ &\;=\; \frac{2(-1)^{n+1}}{n}. \end{aligned} The left side of Parseval is \frac1\pi \int_{-\pi}^{\pi} x^2 \dd x = \frac{2\pi^2}{3}, so \frac{2\pi^2}{3} \;=\; \sum_{n=1}^{\infty} \frac{4}{n^2}, and dividing by 4 gives the result.

NoteEuler’s original argument

Euler compared the Taylor series of \frac{\sin x}{x} = 1 - \frac{x^2}{6} + \cdots with the product \prod_{n\ge1}\left(1 - \frac{x^2}{n^2\pi^2}\right) over its roots, and matched the x^2 coefficients. The product formula was not fully justified until Weierstrass, more than a century later, but the answer was right.

The same trick gives more sums

Using f(x) = x^2 in Parseval gives \zeta(4) = \sum n^{-4} = \pi^4/90, and higher even powers of x give every \zeta(2k) as a rational multiple of \pi^{2k}. The odd values are different. Apéry proved in 1978 that \zeta(3) is irrational, but no closed form for it is known.