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How do i evalulate the following infinite product? $$\displaystyle \prod_{n=2}^{\infty} \left(1-\frac{1}{\zeta(n)}\right)$$ Notation: $\zeta(n)$ is Riemann zeta function.

I'm interested to evaluate the above product. As per the wolfram alpha check the infinite product is approximately going to $0$ but not $0$ which doubt me that it may have certain closed form.

I tried in the following manner.

Let the sequence for $n\geq 1$ be $\zeta_n =1-\frac{1}{\zeta(n)}$ and we are supposed to find the $P=\displaystyle \prod_{1\leq n} \zeta_n$. Since zeta function is decreasing for all $n\geq 2$ function ie $\zeta(n)>\zeta(n+1)$ which is trivial to prove by definition. Now we note that $$\zeta_{n+1}-\zeta_n=\frac{1}{\zeta(n)}-\frac{1}{\zeta(n+1)}<0$$ which implies $\zeta_{n+1} < \zeta_{n} $ and shows our sequence is decreasing sequence. Hence $\operatorname{sup}\left\{ \zeta_n: n\in\mathbb N\right\}=1$ and $\operatorname{inf}\left\{\zeta_n: n\in\mathbb N\right\}=0$ as $\displaystyle \lim_{n\to \infty}\frac{1}{\zeta(n)}=1$ and thus we have bound $0< P <1$.

I'm stuck here. I wish to know if the product has any closed form or it is $0$. Thank you.

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    $\begingroup$ I'm not quite sure I understand your question. Isn't most of the factors in your product less than 1/2? Wouldn't that mean the answer is just 0? $\endgroup$
    – Trebor
    Jun 24 '20 at 7:48
  • $\begingroup$ your product needs to start from $n=2$, since $\zeta(1)$ is not defined. $\endgroup$
    – GreginGre
    Jun 24 '20 at 7:53
  • $\begingroup$ Trebor. I see since $ \zeta(2) <2$ and hence $\frac{1}{\zeta(2)} > \frac{1}{2}$ a nd hence $1-\frac{1}{\zeta(2)} < \frac{1}{2}$. Thank you for hint. $\endgroup$
    – Naren
    Jun 24 '20 at 8:05
  • $\begingroup$ GreginGre, I have edited the post. Thank you. $\endgroup$
    – Naren
    Jun 24 '20 at 8:06
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Let $P = \prod_{n=2}^\infty (1 - 1/\zeta(n))$. (We skip $n=1$ since $\zeta(1)$ is a pole of $\zeta$.) Then note,

$$\ln(P) = \sum_{n=2}^\infty \ln \left( 1 - \frac{1}{\zeta(n)} \right)$$

$\zeta(n)$, as $n \to \infty$ along the positive integers (greater than one), is clearly a positive, monotone decreasing sequence, bounded below by its limit of $1$. Thus, each logarithm has an argument which is slightly less than one, i.e. $1 - 1/\zeta(n) < 1$. Thus, the summands are each less than $\log(1) = 0$. Moreover, $1 - 1/\zeta(n) \to 0$ since $\zeta(n) \to 1$. Thus, $\log(1 - 1/\zeta(n)) \to -\infty$. Then clearly the sum, too, is $-\infty$.

Thus,

$$P = e^{\ln(P)} = 0$$

and thus, your product is zero.

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  • $\begingroup$ Thank you for the solution :) . I even tried taking logarithmic but I didn't see the sum is going to $-\infty$ rather I notice that the sum is going to $+\infty$. $\endgroup$
    – Naren
    Jun 24 '20 at 8:09

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