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I'm interested in the following limit: $$ L = \lim_{N \to \infty} \sum_{a_1 = 1}^\infty \sum_{a_2 = 1}^\infty \cdots \sum_{a_N = 1}^\infty \frac{1}{a_1^2 \, a_2 ^3 \cdots a_N^{N+1}}$$ It can be rewritten as $$ L = \lim_{N \to \infty} \prod_{n = 2}^{N} \zeta (n).$$ or taking log of both sides $$ \ln L = \lim_{N \to \infty} \sum_{n = 2}^{N} \ln \zeta (n). \tag{1}$$ For a long time I couldn't prove/disprove the convergence of $(1)$. First thing to note is that the partial sums are monotonically increasing. My original idea was to prove the upper bound $$ n^2 \,(\zeta(n)-1) < C < \infty \,\,, \forall n \geq 2 \tag{2}$$ Then by some Taylor series manipulation the proof would be done. Question:

$\star$ How could $(2)$ be proven? Mathematica gives that the product $(2)$ should go to zero, but how to prove it mathematically?

$\star$ Or maybe there is a more elegant way to test convergence of $(1)$?

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  • $\begingroup$ $1<\zeta(s)<1/(1-2^{1-s})$ for $s>1$, so $$\sum_{n=2}^\infty\ln\zeta(n)>-\sum_{n=2}^\infty\ln(1-2^{1-s})$$and it is easy to check for convergence from here. $\endgroup$ – Simply Beautiful Art Dec 30 '17 at 15:26
  • $\begingroup$ You meant: $\ln \zeta(n) < -\ln(1 - 2^{1 - n})$, I think? $\endgroup$ – Raito Dec 30 '17 at 15:32
  • $\begingroup$ @Raito Oops, you are right. Though it's too late to edit. $\endgroup$ – Simply Beautiful Art Dec 30 '17 at 15:33
  • $\begingroup$ The product definitely does not go to zero. Trivially,$$\prod_{n=2}^N\zeta(n)\ge\zeta(2)$$Since every term in the product is greater than $1$. $\endgroup$ – Simply Beautiful Art Dec 30 '17 at 15:40
  • $\begingroup$ I meant the product of $n^2$ and $\zeta(n)-1$ which I want to bound. $\endgroup$ – DrLRX Dec 30 '17 at 15:49
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As an alternative way, notice that for $s \geq 2$ we have

$$ 1 \leq \zeta(s) \leq 1 + \frac{1}{2^s} + \int_{2}^{\infty} \frac{dx}{x^s} \leq 1 + \frac{3}{2^s}. $$

So it follows that

$$ \sum_{s=2}^{\infty} \log \zeta(s) \leq \sum_{s=2}^{\infty} \frac{3}{2^s} < \infty $$

and hence the sum converges by the comparison test. Exponentiating shows that the product also converges.

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  • $\begingroup$ I accepted this answer because it doesn't use fancy zeta function identities to prove it. Thank you! :) $\endgroup$ – DrLRX Dec 30 '17 at 19:28
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We have the following relationship between the Riemann zeta and the Dirichlet eta functions:

$$\zeta(s)=\frac{\eta(s)}{1-2^{1-s}}$$

where

$$\eta(s)=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\dots<1,\quad\Re(s)>0$$

Thus, for $s>1$, we have

$$\zeta(s)<\frac1{1-2^{1-s}}$$

and trivially $1<\zeta(s)$.

For any $1<x<y$, we also have $\ln(x)<\ln(y)$, and so

$$\ln\zeta(n)<-\ln(1-2^{1-n})$$

which converges by the ratio test.

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  • $\begingroup$ We have $\frac{1}{2}<\eta(s)<1$ for $s>0$. $\endgroup$ – Wojowu Dec 30 '17 at 15:43
  • $\begingroup$ Many typos! x.x luckily they're just typos I think. $\endgroup$ – Simply Beautiful Art Dec 30 '17 at 15:46
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Given the prime zeta function $P(s)=\sum_{p}\frac{1}{p^s}$ we have $$\log\zeta(s)=\sum_{n\geq 1}\frac{1}{n}P(ns) = \sum_{n\geq 1}\sum_{p}\frac{1}{n p^{ns}} $$ hence $$ \sum_{s\geq 2}\log\zeta(s) =\sum_{n\geq 1}\sum_p \frac{1}{np^n(p^n-1)}$$ and $$ \prod_{s\geq 2}\zeta(s) = \exp\sum_{n\geq 1}\sum_p \frac{1}{np^n(p^n-1)}$$ is clearly convergent.

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