Euler Product Formula

Summary: §270–§277, §283–§284. The identity ${\sum }_{k=1}^{\infty }1/k^n={\prod }_p(1-1/p^n{)}^{-1}$ — a sum over the integers equals a product over the primes. Euler gives two independent derivations: a direct expansion of the reciprocal product using the geometric series and unique factorisation (Movement 2 of Chapter 15), and an Eratosthenes-style sieve of the series itself (§283).

Sources: chapter15

Last updated: 2026-05-11

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The identity

For every $n$ for which the sum converges,

${\sum }_{k=1}^{\infty }\frac{1}{k^n}={\prod }_{p\text{ prime}}\frac{1}{1-1/p^n}={\prod }_{p\text{ prime}}\frac{p^n}{p^n-1}.$

In Euler’s notation (§274), with $P$ for the product and the series fully written out:

$P=\frac{1}{(1-\frac{1}{2^n})(1-\frac{1}{3^n})(1-\frac{1}{5^n})(1-\frac{1}{7^n})\cdots }=1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{5^n}+\cdots .$

Euler notes that “all natural numbers occur with no exception” in the denominators (source: chapter15, §274) — the formula is essentially a restatement of unique factorisation.

Derivation I — Expansion of the reciprocal product (§270–§274)

Given any factors $\alpha ,\beta ,\gamma ,\delta ,\dots$, the geometric-series expansion of each $1/(1-{\alpha }_{i}z)$ followed by multiplication gives

$\frac{1}{(1-\alpha z)(1-\beta z)(1-\gamma z)\cdots }=1+Az+B{z}^2+C{z}^3+\cdots ,$

where $A$ is the sum of the ${\alpha }_i$, $B$ the sum of products taken two at a time with repetition allowed, $C$ three at a time with repetition, etc. (source: chapter15, §270; contrast with §264, where the linear factors $1+\alpha z$ produce only distinct products).

Setting $z=1$ and ${\alpha }_i=1/p_i^n$ over the primes (§274), each entry in the expanded series is

$\frac{1}{p_1^{n{e}_1}{p}_2^{n{e}_2}{p}_3^{n{e}_3}\cdots }=\frac{1}{(p_1^{e_1}{p}_2^{e_2}\cdots {)}^n}=\frac{1}{k^n}$

for some natural number $k$ with prime factorisation $p_1^{e_1}{p}_2^{e_2}\cdots$. By unique factorisation each natural number appears exactly once as such a product, giving the formula.

The instructive intermediate case (§272) is ${\alpha }_1=1/2$ alone: $1/(1-1/2)=1+1/2+1/4+1/8+\cdots$, the powers of $1/2$. Adding ${\alpha }_2=1/3$ gives powers of $1/2$ times powers of $1/3$ — the $\{2,3\}$-smooth numbers. Including all primes covers every natural number.

Derivation II — Eratosthenes sieve on the series (§283)

Independently and without reference to the product expansion, start from

$A=1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{5^n}+\cdots .$

Subtracting $A/2^n=1/2^n+1/4^n+1/6^n+\cdots$ removes every even-index term:

$B=\left(1-\frac{1}{2^n}\right)A=1+\frac{1}{3^n}+\frac{1}{5^n}+\frac{1}{7^n}+\frac{1}{9^n}+\cdots .$

Subtracting $B/3^n=1/3^n+1/9^n+1/15^n+\cdots$ removes every term whose index is divisible by 3:

$C=\left(1-\frac{1}{3^n}\right)B=1+\frac{1}{5^n}+\frac{1}{7^n}+\frac{1}{11^n}+\frac{1}{13^n}+\cdots .$

Continuing with primes $5,7,11,\dots$, each step kills all multiples of one further prime. After sieving by every prime only the term 1 survives, so

$A{\prod }_{p\text{ prime}}\left(1-\frac{1}{p^n}\right)=1,$

equivalent to the formula above.

This is the Sieve of Eratosthenes lifted to the series level: where Eratosthenes erases multiples of $p$ from a list of integers, Euler subtracts a scaled copy of the running series.

The Möbius dual (§275)

Pairing the formula with the §269 product $\prod (1-1/p^n)=1-1/2^n-1/3^n-1/5^n+1/6^n-\cdots$ (Möbius signs over squarefree integers; see squarefree-and-mobius-series) gives the reciprocal relation

$P\cdot Q=1\text{with}P={\sum }_k\frac{1}{k^n},Q={\sum }_k\frac{\mu (k)}{k^n}.$

Symbolically, $1/\zeta (n)=\sum \mu (k)/k^n$.

The alternating variant (§284)

Applying the §283 sieve to the alternating series

$A=1-\frac{1}{3^n}+\frac{1}{5^n}-\frac{1}{7^n}+\frac{1}{9^n}-\cdots$

(no even terms, since only odd-index terms appear from the start; signs follow the $4m+1$ vs $4m-1$ rule) produces

$1=A{\prod }_{p\equiv 1\mathrm{mod}4}(1-\frac{1}{p^n}){\prod }_{p\equiv -1\mathrm{mod}4}(1+\frac{1}{p^n}),$

i.e. $A={\prod }_p(1-{\chi }_4(p)/p^n{)}^{-1}$ in modern notation, where ${\chi }_4$ is the non-trivial Dirichlet character mod 4. See prime-sign-series-for-pi for the $\pi$-series corollaries that fall out at $n=1,3$.

Worked first values

$n$	$\sum 1/k^n$	Product over primes
1	$\infty$	${\prod }_p\frac{p}{p-1}$ (diverges; see divergence-of-prime-reciprocals)
2	${\pi }^2/6$	${\prod }_p\frac{p^2}{p^2-1}=\frac{4}{3}\cdot \frac{9}{8}\cdot \frac{25}{24}\cdot \frac{49}{48}\cdots$
4	${\pi }^4/90$	${\prod }_p\frac{p^4}{p^4-1}$
6	${\pi }^6/945$	${\prod }_p\frac{p^6}{p^6-1}$

The right column for $n=2$ appears in Euler’s “Example I” at §277.

Significance

The Euler product formula is the seed of analytic number theory:

• Equating an additive sum (over integers) with a multiplicative product (over primes) lets one trade information about integers for information about primes.
• Logarithmic differentiation gives $\log \zeta (s)={\sum }_p{\sum }_{k}1/(k{p}^{ks})$, which separates $\sum 1/p^s$ from the higher prime-power tail (§278).
• At $s=1$ this yields the divergence of $\sum 1/p$ (§279), a quantitative refinement of Euclid’s theorem.
• Dirichlet (1837) generalised the formula by replacing $1/p^s$ with $\chi (p)/p^s$ for a character $\chi$, recovering Dirichlet’s theorem on primes in arithmetic progressions. Euler had already noticed the relevant character series — the §284, §294, §295 variants below — without the general framework.
• Riemann (1859) analytically continued $\zeta (s)$ to $\mathbb{C}\setminus \{1\}$; the Euler product becomes the bridge between the zeros of $\zeta$ and the distribution of primes.

Related pages

• chapter-15-on-series-which-arise-from-products
• squarefree-and-mobius-series
• divergence-of-prime-reciprocals
• prime-zeta-values
• prime-sign-series-for-pi
• basel-problem
• zeta-at-even-integers
• arctangent-series
• geometric-series
• logarithmic-series