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[parent] Euler product formula (Theorem)

Theorem (Euler). If $s > 1$ , the infinite product

$\displaystyle \prod_{p}\frac{1}{1-\frac{1}{p^s}}$ (1)

where $p$ runs the positive rational primes, converges to the sum of the over-harmonic series
$\displaystyle \sum_{n=1}^\infty\frac{1}{n^s} \,=\, \zeta(s).$ (2)

Proof. Denote the sequence of prime numbers by $p_1 < p_2 < p_3 <\,\ldots$ For any $s > 0$ , we can form convergent geometric series $$\frac{1}{1-\frac{1}{p_1^s}} \,=\, 1+\frac{1}{p_1^s}+\frac{1}{p_1^{2s}}+\ldots \,=\, \sum_{\nu_1=0}^\infty\frac{1}{p_1^{\nu_1s}},$$ $$\frac{1}{1-\frac{1}{p_2^s}} \,=\, 1+\frac{1}{p_2^s}+\frac{1}{p_2^{2s}}+\ldots \,=\, \sum_{\nu_2=0}^\infty\frac{1}{p_2^{\nu_2s}}.$$ Since these series are absolutely convergent, their product (see multiplication of series) may be written as $$\frac{1}{1-\frac{1}{p_1^s}}\cdot\frac{1}{1-\frac{1}{p_2^s}} \;=\; \sum_{\nu_1,\nu_2=0}^\infty\frac{1}{p_1^{\nu_1s}}\cdot\frac{1}{p_2^{\nu_2s}} \;=\; \sum_{\nu_1,\nu_2=0}^\infty\frac{1}{\left(p_1^{\nu_1}p_2^{\nu_2}\right)^s}$$ where $\nu_1$ and $\nu_2$ independently on each other run all nonnegative integers. This equation can be generalised by induction to

$\displaystyle \prod_{\nu=1}^k \frac{1}{1-\frac{1}{p_\nu^s}} \;=\; \sum_{\nu_1,\... ...nu_k=0}^\infty\frac{1}{\left(p_1^{\nu_1}p_2^{\nu_2}\cdots p_k^{\nu_k}\right)^s}$ (3)

for $s > 0$ and for arbitrarily great $k$ ; the exponents $\nu_1,\,\nu_2,\,\ldots,\,\nu_k$ run independently all nonnegative integers.

Because the prime factorization of positive integers is unique, we can rewrite (3) as

$\displaystyle \prod_{\nu=1}^k \frac{1}{1-\frac{1}{p_\nu^s}} \;=\; \sum_{(n)}\frac{1}{n^s},$ (4)

where $n$ runs all positive integers not containing greater prime factors than $p_k$ . Then the inequality
$\displaystyle \sum_{n=1}^{p_k}\frac{1}{n^s} < \prod_{\nu=1}^k \frac{1}{1-\frac{1}{p_\nu^s}},$ (5)

holds for every $k$ , since all the terms $1,\,\frac{1}{p_1^s},\,\ldots,\,\frac{1}{p_k^s}$ are in the series of the right hand side of (4). On the other hand, this series contains only a part of the terms of (2). Thus, for $s > 1$ , the product (3) is less than the sum $\zeta(s)$ of the series (2), and consequently
$\displaystyle \sum_{n=1}^{p_k}\frac{1}{n^s} < \prod_{\nu=1}^k \frac{1}{1-\frac{1}{p_\nu^s}} < \zeta(s).$ (6)

Letting $k \to \infty$ , we have $p_k \to \infty$ , and the sum on the left hand side of (6) tends to the limit $\zeta(s)$ , therefore also tends the product (3). Hence we get the limit equation
$\displaystyle \prod_{p}\frac{1}{1-\frac{1}{p^s}} \;=\; \zeta(s) \qquad (s > 1).$ (7)

Bibliography

1
E. LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset III.2. Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1940).




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See Also: Riemann zeta function, Euler product


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Cross-references: limit, left hand side, contains, right hand side, inequality, prime factors, prime factorization, exponents, induction, equation, integers, multiplication of series, product, absolutely convergent, series, geometric series, convergent, sequence, proof, over-harmonic series, sum, converges, rational primes, positive, infinite product, Euler, theorem
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This is version 3 of Euler product formula, born on 2008-12-28, modified 2008-12-29.
Object id is 11404, canonical name is EulerProductFormula2.
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Classification:
AMS MSC11A41 (Number theory :: Elementary number theory :: Primes)
 11A51 (Number theory :: Elementary number theory :: Factorization; primality)
 11M06 (Number theory :: Zeta and $L$-functions: analytic theory :: $\zeta $)
 40A20 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of infinite products)

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