Is there any closed form for $\displaystyle \prod_{n=1}^{\infty} \left(1-\frac{1}{\zeta(n)}\right)$

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.

• 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? Jun 24 '20 at 7:48
• your product needs to start from $n=2$, since $\zeta(1)$ is not defined. Jun 24 '20 at 7:53
• 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. Jun 24 '20 at 8:05
• GreginGre, I have edited the post. Thank you. Jun 24 '20 at 8:06

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.

• 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 $+\infty$. Jun 24 '20 at 8:09