Basel Problem

Summary: §167: Euler’s celebrated evaluation ${\sum }_{k=1}^{\infty }1/k^2={\pi }^2/6$, obtained by combining the sinh infinite product with Newton’s identities. The first solved member of an infinite family of even-zeta values (see zeta-at-even-integers).

Sources: chapter10, chapter15

Last updated: 2026-05-11

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Statement

$\boxed{1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots =\frac{{\pi }^2}{6}}$

(source: chapter10, §167). The problem of evaluating this sum was posed by Pietro Mengoli in 1644 and resisted Jacob Bernoulli, Johann Bernoulli, Leibniz, and de Moivre for nearly a century, becoming famous as the Basel problem after the city where the Bernoullis worked.

Euler’s derivation

Start from the §156 product

$\frac{e^x-e^{-x}}{2}=x\left(1+\frac{x^2}{1\cdot 2\cdot 3}+\frac{x^4}{1\cdot 2\cdot 3\cdot 4\cdot 5}+\frac{x^6}{1\cdot 2\cdots 7}+\cdots \right)=x{\prod }_{k=1}^{\infty }\left(1+\frac{x^2}{k^2{\pi }^2}\right).$

Divide by $x$ and substitute $x^2={\pi }^{2}z$:

$1+\frac{{\pi }^{2}z}{1\cdot 2\cdot 3}+\frac{{\pi }^4{z}^2}{1\cdot 2\cdot 3\cdot 4\cdot 5}+\frac{{\pi }^6{z}^3}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7}+\cdots =(1+z)\left(1+\frac{z}{4}\right)\left(1+\frac{z}{9}\right)\left(1+\frac{z}{16}\right)\cdots .$

Now apply the §165 framework. The series side has

$A=\frac{{\pi }^2}{6},B=\frac{{\pi }^4}{120},C=\frac{{\pi }^6}{5040},D=\frac{{\pi }^8}{362880},\dots$

The product side has roots ${\alpha }_k=1/k^2$ for $k=1,2,3,\dots$. By §166,

$P={\alpha }_1+{\alpha }_2+{\alpha }_3+\cdots =1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots =A=\frac{{\pi }^2}{6}.$

Done.

Why this is striking

A real-numbers sum, with no apparent connection to geometry, equals ${\pi }^2/6$. The appearance of $\pi$ — a number whose definition is geometric — in the answer to a purely arithmetic question is the kind of cross-domain coincidence that suggests deep structure.

Euler’s resolution: $\pi$ appears because the zeros of $\sin z$ are at $z=k\pi$. The series $\sum 1/k^2$ is encoded in the coefficients of the power series of $\sin z$ when read against the zeros of $\sin z$ via the product expansion. The ${\pi }^2/6$ is, finally, a logarithmic derivative phenomenon — Euler is computing the trace of an operator whose spectrum is geometrically determined.

A second derivation: the sine product directly

Equivalent to the above, but more direct. From the sine product

$\sin z=z{\prod }_{k=1}^{\infty }\left(1-\frac{z^2}{k^2{\pi }^2}\right)=z-z\cdot {\sum }_{k=1}^{\infty }\frac{z^2}{k^2{\pi }^2}+(\text{higher})$

and the power series

$\sin z=z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots ,$

equate coefficients of $z^3$:

$-\frac{1}{6}=-{\sum }_{k=1}^{\infty }\frac{1}{k^2{\pi }^2}⟹{\sum }_{k=1}^{\infty }\frac{1}{k^2}=\frac{{\pi }^2}{6}.$

Euler does not state this version explicitly in §167 (he goes through the sinh product and Newton’s identities), but it is the same calculation with the imaginary substitution $x\mapsto iz$ already applied; see sine-infinite-product for the direct one-line argument.

Higher powers

The same Newton recurrence yields all even-power sums:

$Q={\sum }_{k=1}^{\infty }\frac{1}{k^4}=\frac{{\pi }^4}{90},R={\sum }_{k=1}^{\infty }\frac{1}{k^6}=\frac{{\pi }^6}{945},S={\sum }_{k=1}^{\infty }\frac{1}{k^8}=\frac{{\pi }^8}{9450},$

$T={\sum }_{k=1}^{\infty }\frac{1}{k^{10}}=\frac{{\pi }^{10}}{93555},\dots$

(source: chapter10, §167). See zeta-at-even-integers for the systematic table.

Re-derivation via primes (Chapter 15)

Chapter 15 gives a second route into $\zeta (2)={\pi }^2/6$: the Euler product

$\frac{{\pi }^2}{6}={\prod }_{p\text{ prime}}\frac{p^2}{p^2-1}=\frac{4}{3}\cdot \frac{9}{8}\cdot \frac{25}{24}\cdot \frac{49}{48}\cdot \frac{121}{120}\cdots$

Combined with the Wallis product, this yields the §285 catalogue of identities for $\pi /2$, $\pi /4$, etc. as prime-only ratios. Logarithmically, the formula feeds the §278 transposition that produces the [[divergence-of-prime-reciprocals|divergence of $\sum 1/p$]].

Modern footnote

In modern notation $\sum 1/k^2=\zeta (2)$, where $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ is the Riemann zeta function. Euler’s result is the first nontrivial value: $\zeta (2)={\pi }^2/6$. He later evaluates $\zeta (2k)$ for every positive integer $k$ — see zeta-at-even-integers — but $\zeta (2k+1)$ for odd indices remains, in 2026 as in Euler’s day, almost completely opaque. Apéry showed in 1978 that $\zeta (3)$ is irrational; nothing comparable is known for $\zeta (5),\zeta (7),\dots$.

The lacuna at odd indices is precisely the gap left by Euler’s method: his factorization argument applies to functions whose zeros generate a geometric progression in arclength (e.g. $\sin z$ at $k\pi$), and there is no comparable “closed-form” function whose squared-zero sum is $\zeta (3)$.

Related pages

• zeta-at-even-integers
• odd-and-alternating-zeta-decomposition
• newtons-identities
• sine-infinite-product
• exponential-infinite-product
• sine-and-cosine-series
• pi
• chapter-9-on-trinomial-factors
• chapter-10-on-the-use-of-the-discovered-factors-to-sum-infinite-series
• chapter-15-on-series-which-arise-from-products
• euler-product-formula
• divergence-of-prime-reciprocals
• prime-sign-series-for-pi