Virtual laboratory/Mathematics/Calculus/The Basel Problem — π in the Sum of Reciprocal Squares

The Basel Problem — π in the Sum of Reciprocal Squares

An extension of the topic of series and limits. The sum of reciprocal squares 1+1/4+1/9+1/16+… is convergent (unlike the divergent harmonic series) and — surprisingly — equals exactly π²/6≈1.645. The appearance of π in a purely arithmetic sum is one of the most beautiful results in mathematics. The problem was posed by Mengoli (1644) and solved by the 28-year-old Leonhard Euler in 1735.

The Basel Problem — π in the Sum of Reciprocal Squares
Mathematics · Calculus · Basel problem — π in the sum of reciprocal squares (1735)
Number of terms NN10
The sum 1+14+19+116+…1+\tfrac14+\tfrac19+\tfrac1{16}+\dots (to infinity) is…
terms N
10
last term 1/N²
0.010000
sum S(N)
1.54977
limit π²/6
1.64493
π²/6 − S(N)
0.09517
🎯 Challenge: hit the sum S(N)
streak: 0

I will give you a target partial sum — use the slider to pick the number of terms N until S(N) gets close to it. The limit is π²/6 ≈ 1.6449.

What just happened?

You were adding successive reciprocal squares: 1, 14, 19, 116,…1,\ \tfrac14,\ \tfrac19,\ \tfrac1{16},\dots The successive terms shrink quadratically, so the partial sum very quickly stops growing appreciably and settles just below a certain ceiling. That ceiling is exactly π2/6\pi^2/6 — and that is the biggest surprise: the number π\pi, tied to the circle, appears in a purely arithmetic sum.

The Basel problem:

1+14+19+116+⋯=π261+\tfrac14+\tfrac19+\tfrac1{16}+\dots=\dfrac{\pi^2}{6}

The task is to find the exact sum of the series of reciprocal squares. It was posed by Pietro Mengoli in 1644, and for nearly 90 years the greatest mathematicians (among them the Bernoullis) tried to solve it, without success. The name comes from the Swiss city of Basel, home of the Bernoulli family who worked on it.

Euler's solution (1735):

The problem was finally solved by the 28-year-old Leonhard Euler, who proved an astonishing result: the sum equals exactly π2/6≈1.6449\pi^2/6 \approx 1.6449. This achievement brought him immediate fame. The result also launched the study of the Riemann zeta function: in its language ζ(2)=∑n=1∞1n2=π26\zeta(2)=\sum_{n=1}^{\infty}\tfrac1{n^2}=\tfrac{\pi^2}{6} — one of the pillars of number theory.

Convergent, not divergent — why?

Compare with the harmonic series 1+12+13+…1+\tfrac12+\tfrac13+\dots, which is divergent (its sum tends to infinity). There the terms 1/n1/n shrink too slowly. Here the terms 1/n21/n^2 shrink much faster — quadratically — and that is enough for the sum to be finite. The series ∑1/n2\sum 1/n^2 is therefore convergent, and its limit is precisely π2/6\pi^2/6.

Why this matters on the exam:

This is the classic pair to remember: ∑1/n\sum 1/n — divergent, ∑1/n2\sum 1/n^2 — convergent. The rate at which the terms shrink decides whether the series converges. The condition an→0a_n\to 0 alone is not enough — what matters is how fast they shrink.

Remember: the series ∑n=1∞1/n2\sum_{n=1}^{\infty} 1/n^2 is convergent, and its sum equals exactly π2/6≈1.6449\pi^2/6 \approx 1.6449. The surprising appearance of π\pi in the sum of reciprocal squares is one of the most beautiful results in mathematics (Euler, 1735).

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Theory & formulas

The sum of reciprocal squares

∑n=1∞1n2=1+14+19+116+⋯=π26≈1.645\sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac14+\frac19+\frac1{16}+\cdots=\frac{\pi^2}{6}\approx 1.645

Why it converges

The terms 1n2\tfrac1{n^2} shrink quadratically — much faster than 1n\tfrac1n in the harmonic series — so the sum is finite.

The surprise

In a purely arithmetic sum the constant π\pi appears — a constant from the geometry of the circle. The result was found by Euler in 1735.

Notation: nn the index of a term, π≈3.14159\pi\approx 3.14159.

Frequently asked questions

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What is the value of the sum 1+1/4+1/9+…?

Exactly π²/6 ≈ 1.645 — the series is convergent.

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Why is this series convergent while the harmonic series is not?

Because the terms 1/n² shrink much faster (quadratically) than the terms 1/n of the harmonic series.

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Who solved the Basel problem?

Leonhard Euler in 1735, at the age of 28 — which brought him immediate fame.

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