Prove-that-s-prime-1-1-p-s-

Question Number 142875 by Snail last updated on 06/Jun/21

P r o v e t h a t ζ (s) = \prod_{p r i m e} \frac{1}{1 - p^{- s}}

Answered by Dwaipayan Shikari last updated on 06/Jun/21

ζ (s) = 1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + \dots

\frac{ζ (s)}{2^{s}} = \frac{1}{2^{s}} + \frac{1}{4^{s}} + \frac{1}{6^{s}} + \dots (M u l t i p l e s o f 2)

ζ (s) (1 - \frac{1}{2^{s}}) = 1 + \frac{1}{3^{s}} + \frac{1}{5^{s}} + \frac{1}{7^{s}} + \dots (S u b t r a c t)

ζ (s) (1 - \frac{1}{2^{s}}) \frac{1}{3^{s}} = \frac{1}{3^{s}} + \frac{1}{9^{s}} + \frac{1}{15^{s}} + \frac{1}{21^{s}} + \dots (a l l m u l t i p l e s o f 3)

ζ (s) (1 - \frac{1}{2^{s}}) - ζ (s) (1 - \frac{1}{2^{s}}) \frac{1}{3^{s}} = \frac{1}{1} + \frac{1}{5^{s}} + \frac{1}{7^{s}} + . .

ζ (s) (1 - \frac{1}{2^{s}}) (1 - \frac{1}{3^{s}}) = 1 + \frac{1}{5^{s}} + \frac{1}{7^{s}} + . .

S i n c e e a c h n u m b e r s (\neq 1) a r e b u l i d f r o m p r i m e s,

w e c a n k e e p g o i n g l i k e t h i s

ζ (s) (1 - \frac{1}{2^{s}}) (1 - \frac{1}{3^{s}}) (1 - \frac{1}{5^{s}}) . . = 1

ζ (s) = \underset{p}{\prod^{\infty}} \frac{1}{1 - \frac{1}{p^{s}}}

Commented by Snail last updated on 06/Jun/21

T h a n k s

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