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I am quite certain that this limit:

$$\lim_{s\to 1} \, \zeta (s) \prod _{n=1}^k \left(1-\frac{1}{n^{s-1}}\right)=0$$ is always zero for any integer $k \geq 2$.

Can you prove it? I can't.

I have to go outdoors now.

As a Mathematica program this is:

Table[Limit[Zeta[s]*Product[(1 - 1/n^(s - 1)), {n, 1, k}], s -> 1], {k, 2, 12}]
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The factor with $n=1$ is constantly zero. – Hagen von Eitzen Jul 21 '13 at 12:03
If $n$ starts from 2 and $k>n$ then the limit is $0$. – Mhenni Benghorbal Jul 21 '13 at 12:13
up vote 3 down vote accepted

From $1-\frac1{1^{s-1}}=0$, the expression is zero for any $s\ne1$, hence also zero in the limit

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Try it with the product starting at 2 instead of 1. The problem as stated has this obvious error. – marty cohen Jul 22 '13 at 3:15

To make things a bit more interesting, I will assume that you intended to start the product from $n=2$.

In this answer, I use the fact that $$ \zeta(z)(1-2^{1-z})=1^{-z}-2^{-z}+3^{-z}-4^{-z}+\dots $$ The idea there is to use the fact that as $z\to1$ the limit of the right hand side is $\log(2)$. Thus, we have $$ \begin{align} \lim_{z\to1}\zeta(z)\prod_{n=2}^k(1-n^{1-z}) &=\lim_{z\to1}\zeta(z)(1-2^{1-z})\prod_{n=3}^k(1-n^{1-z})\\ &=\log(2)\prod_{n=3}^k\lim_{z\to1}(1-n^{1-z})\\ &=0 \end{align} $$ for $k\ge3$ since $\lim\limits_{z\to1}(1-n^{1-z})=0$ for each $n$.

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Yes, you are correct it was my intention: – Mats Granvik Aug 2 '13 at 19:28
The second Mathematica program at this link is related to this question: – Mats Granvik Aug 2 '13 at 19:42

$1-a^{1-s} = 1 - e^{(1-s)\ln a} = (s-1)\ln a + \mathcal{O}((1-s)^2)$. that's how we can prove (*) that $\lim_{s \to 1} (1-2^{1-s}) \zeta(s) = \ln 2$.

now $1-a^{1-s}$ has a simple zero at $s=1$, thus $\prod_{n=2}^k (1-n^{1-s})$ has a $0$ of order $k-1$ at $s=1$, and $\lim_{s \to 1} \zeta(s)^{k-1} \prod_{n=2}^k (1-n^{1-s}) = \prod_{n=2}^k \ln n$ .

(*) and the fact that $\zeta(s) = \frac{s}{s-1}- s \int_1^\infty \{ x \} x^{-s-1}dx $

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