A partial Euler product is a finite truncation of an Euler product, obtained by restricting the product to primes up to a specified bound. For the Riemann zeta function, it takes the form
P ( s , x ) = ∏ p ≤ x ( 1 − 1 p s ) − 1 , {\displaystyle P(s,x)=\prod _{p\leq x}\left(1-{\frac {1}{p^{s}}}\right)^{-1},}
where the product is over primes p ≤ x {\displaystyle p\leq x} . As x {\displaystyle x} increases, the partial product approaches the full Euler product in its region of convergence. Analogous partial products can be formed for Dirichlet L-functions and other Dirichlet series with Euler products. Taking logarithms of partial Euler products expresses them as sums over prime powers and allows their asymptotic behaviour to be studied. A classical result concerning partial Euler products is Mertens' third theorem, which gives an asymptotic formula for the product over primes of ( 1 − 1 / p ) − 1 {\displaystyle (1-1/p)^{-1}} . Partial Euler products also arise in the study of elliptic curve L-functions, including work related to the Birch and Swinnerton-Dyer conjecture and the Riemann hypothesis for such L-functions.
Overview In a branch of mathematics called analytic number theory, numbers such as prime numbers are studied using "smooth" (continuous) tools from other areas. One of those tools are Euler Products; in short, if you have a sum like 1 + 1 4 + 1 9 + 1 16 + 1 25 ⋯ {\displaystyle 1+{\frac {1}{4}}+{\frac {1}{9}}+{\frac {1}{16}}+{\frac {1}{25}}\cdots }
(which is the sum of the reciprocals of squares, so ζ ( 2 ) {\displaystyle \zeta (2)} , the Riemann Zeta function), then instead of manually adding every term, we can rely on the fact that every composite number has a prime factorization, doing some analysis we get:
ζ ( s ) = ∏ p ( 1 − 1 p s ) − 1 {\displaystyle \zeta (s)=\prod _{p}\left(1-{\frac {1}{p^{s}}}\right)^{-1}}
For ζ ( 2 ) , {\displaystyle \zeta (2),} this gives us:
( 1 − 1 2 2 ) − 1 ( 1 − 1 3 2 ) − 1 ( 1 − 1 5 2 ) − 1 ( 1 − 1 7 2 ) − 1 ( 1 − 1 11 2 ) − 1 ( 1 − 1 13 2 ) − 1 ⋯ {\displaystyle \left(1-{\frac {1}{2^{2}}}\right)^{-1}\left(1-{\frac {1}{3^{2}}}\right)^{-1}\left(1-{\frac {1}{5^{2}}}\right)^{-1}\left(1-{\frac {1}{7^{2}}}\right)^{-1}\left(1-{\frac {1}{11^{2}}}\right)^{-1}\left(1-{\frac {1}{13^{2}}}\right)^{-1}\cdots }
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