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Let $p_1, ... , p_n, ...$ be the prime numbers in order. Let $n \in \mathbb{N}$ and $q_1, ..., q_n \in \mathbb{N}$. Define $$ P_n = \prod_{k=1}^n p_k^{q_k} \hspace{1cm} Q_n = \prod_{k=1}^n \left( p_k^{q_k} + u_k \right) $$ where $u_k \in \{-1,+1\}$. Can we prove that if $q_k \geq q$ $$ \frac{P_n}{Q_n} \to^{n\to \infty, q \to \infty} 1 \hspace{1cm}? $$

For instance if $q_1 = ... = q_n$ and $u_k = -1$ then $$ \frac{P_n}{Q_n} = \prod_{k=1}^n \frac{1}{1 - p_k^{-q}} \to^{n \to \infty} \zeta(q) \to^{q \to \infty} 1 $$ Can some similar result be obtained for the slightly more general case presented above?

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  • $\begingroup$ Do you know what happens in the case when all the $q_k$ are equal to $q$ but all the $u_k$ are $+1$? $\endgroup$
    – AnCar
    Commented May 22 at 20:34
  • $\begingroup$ the term in the product is $\frac{p_k^q}{1+p_k^q} = 1 - \frac{1}{p_k^q+1}$ but I do not know where to go from there ... $\endgroup$
    – C Marius
    Commented May 22 at 20:42
  • $\begingroup$ I am not an analytic number theory guy, so my apologies if I am asking something stupid. First, is it correct that Euler product formula for the zeta function holds whenever the real part of the argument is bigger than $1$? Second, is the convergence of zeta at infinity but off the real line known? E.g. is it true that $\zeta(q+i\pi) \to 1$ as well when $q\to\infty$? $\endgroup$
    – AnCar
    Commented May 22 at 20:52
  • $\begingroup$ I do not know that for sure at the moment and is late for me to think too deep. A quick look at the first pic in Wikipedia en.wikipedia.org/wiki/Riemann_zeta_function (the one with domain coloring) seems to agree to $\zeta(q + i\pi) \to 1$ as well. Do you have an idea? $\endgroup$
    – C Marius
    Commented May 22 at 21:03
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    $\begingroup$ Thanks for the chat ... I'll look again tomorrow! It was your work which inspired me! Please have a check your self and write an answer and I'll accept it if it is ok! $\endgroup$
    – C Marius
    Commented May 22 at 21:35

2 Answers 2

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Let's look at the reciprocal $$ \frac{Q_n}{P_n} = \prod_{k=1}^n \biggl( 1 + \frac{u_k}{p^{q_k}} \biggr) $$ for convenience. If $q>1$ and $q_1,\dots,q_n \ge q$, then $$ \frac1{\zeta(q)} = \prod_{k=1}^\infty \biggl( 1 - \frac{1}{p^q} \biggr) \le \frac{Q_n}{P_n} \le \prod_{k=1}^\infty \biggl( 1 + \frac{1}{p^q} \biggr) = \frac{\zeta(q)}{\zeta(2q)}; $$ therefore $\lim_{q\to\infty} \frac{Q_n}{P_n} = 1$ by the squeeze theorem, uniformly in $n$ and thus regardless of whether $n\to\infty$ or not. A similar proof holds whenever $u_k\in\Bbb C$ is uniformly bounded (as long as we avoid individually vanishing factors).

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  • $\begingroup$ Very clean way of writing this by taking the reciprocal from the beginning. $\endgroup$
    – AnCar
    Commented May 22 at 22:16
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Just summarizing the discussion above with the original poster. The solution is virtually theirs.

Claim: Let $p>1$, $a\geq A>0$ such that $p^{-A}\ll 1$ and let $\varepsilon\in\{\pm 1\}$. Then $\tfrac{1}{E(p,A)}\leq \tfrac{1}{1+\varepsilon p^{-a}}\leq E(p,A)$, where $E(p,A)=\tfrac{1}{1-p^{-A}}$.

Proof: This is a simple chain of inequalities. First note that $\tfrac{1}{1+p^{-a}}\leq \tfrac{1}{1+\varepsilon p^{-a}} \leq \tfrac{1}{1-p^{-a}}$.

Now, $\tfrac{1}{\tfrac{1}{1-p^{-A}}}= \tfrac{1}{\tfrac{p^A}{p^A-1}}=1-\tfrac{1}{p^A}\leq 1-\tfrac{1}{p^a}\leq 1-\tfrac{1}{p^a+1}=\tfrac{p^a}{p^a+1}=\tfrac{1}{1+p^{-a}}$, so this proves the left desired inequality.

On the other hand, $\tfrac{1}{1-p^{-a}}\leq \tfrac{1}{1-p^{-A}}$ is equivalent to $1-p^{-A}\leq 1-p^{-a}$, which is equivalent to $p^{-a}\leq p^{-A}$ and $p^A\leq p^a$, which is clearly true. Proof of claim done.

Now, coming back to the problem, it is easy to see that $\tfrac{P_n}{Q_n}=\prod_{k=1}^n \tfrac{1}{1+u_k p_k^{-q_k}}$. We can apply the claim above for each term in the product with $A=q$, $a=q_k$, $p=p_k$ and $\varepsilon=u_k$. This gives us that \begin{equation} \tfrac{1}{\prod_{k=1}^n \tfrac{1}{1-p_k^{-q}}}\leq\tfrac{P_n}{Q_n}\leq \prod_{k=1}^n \tfrac{1}{1-p_k^{-q}} \end{equation}

But note that $\prod_{k=1}^n \tfrac{1}{1-p_k^{-q}}\leq \prod_{k=1}^\infty \tfrac{1}{1-p_k^{-q}}=\zeta(q)$ with the equality being the Euler product formula.

Thus, $\tfrac{1}{\zeta(q)}\leq\tfrac{P_n}{Q_n}\leq \zeta(q)$. As the original poster pointed out, $\lim_{q\to\infty}\zeta(q)=1$ so this proves the desired result by taking the limit as $q\to\infty$.

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