Talk:Proof of the Euler product formula for the Riemann zeta function

Infinite series notation

In the "Proofs" section, the terms of the infinite product are shown with the ellipsis on the left side of the product instead of on the more usual right side. Wouldn't it be a bit clearer (and ncer looking) if the ellipses were moved to the right sides? — Loadmaster (talk) 18:38, 1 January 2011 (UTC)

Correction

I made a small correction to the section titled "The Euler product formula". The original text specified that "1+3" as the numerator for the fraction contained in the series definition of the zeta function. I replaced this with "1". Presumably this is acceptable and if I am in error, please reverse my revision as appropriate. - Letsgoexploring (talk) 00:06, 7 July 2011 (UTC)

First Reference

Apart from a single mention of Euler the first of the "notes" given seems to have no connection with this article, being a discussion of the history of calculus. I suggest it should be removed. Redcliffe maven (talk) 21:33, 1 March 2016 (UTC)

another proof

Let $A_{k}=\{n\in \mathbb {N} ,{\text{ largest prime factor of }}n\leq k\}$. Using the fundamental theorem of arithmetic

$\prod _{p\in {\mathcal {P}},p\leq k}(1+p^{-s}+(p^{2})^{-s}+(p^{3})^{-s}+\ldots )=\sum _{n\in A_{k}}n^{-s}$

For $s>1$ the sequence $\{n^{-s}\}_{n\in \mathbb {N} }$ is non-negative and summable, which means the order of summation doesn't matter, thus

$\zeta (s)=\lim _{N\to \infty }\sum _{n\in [1,N]}n^{-s}=\lim _{k\to \infty }\sum _{n\in A_{k}}n^{-s}$

And for $s\leq 1$, since each term is positive, if $\lim _{k\to \infty }\sum _{n\in A_{k}}n^{-s}$ converges then $\lim _{N\to \infty }\sum _{n\in [1,N]}n^{-s}$ converges, a contradiction since we know it diverges by comparison with $\int _{1}^{N}x^{-s}dx={\frac {N^{1-s}-1}{1-s}}$

78.196.93.135 (talk) 02:58, 18 September 2016 (UTC)

Pseudomathematics should be removed

The section titled The case s = 1 contains mathematical nonsense: pseudomathematics.

For instance, the first two lines read as follows:

"An interesting result can be found for ζ(1), the harmonic series:"

"...(1 - 1/11^s) (1 - 1/7^s) (1 - 1/5^s) (1 - 1/3^s) (1 - 1/2^s) ζ(1) = 1"

"which can also be written as " ... [I have omitted the rest].

I would really like to see any rigorous derivation of that supposed equation above!108.245.209.39 (talk) 08:16, 24 October 2017 (UTC)

I think the section definitely makes sense. It starts with the special case of the first equality of the page for s=1, rearranged, and gives some discussion on it. The general equality has multiple (sketches of) proofs in other sections. 2A02:AB88:3980:2100:D436:8A98:6EE2:9CA8 (talk) 07:23, 20 November 2023 (UTC)

Another proof

Each factor (for a given prime p) in the product above can be expanded to a geometric series consisting of the reciprocal of p raised to multiples of s, as follows...

Can someone add more information about how to do this? The text says "can be expanded" but how it can be expanded? I think more explicit information is needed. Also, which product is referenced with "the product above"? Can we be more specific? Zeyn1 (talk) 14:21, 18 March 2024 (UTC)