This Riemann zeta function limit is always zero for any k, right?

Question

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}]

Answer 1

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

Answer 2

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 $n$.