# Convergence of a product involving primes

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?

• Do you know what happens in the case when all the $q_k$ are equal to $q$ but all the $u_k$ are $+1$? Commented May 22 at 20:34
• 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 ... Commented May 22 at 20:42
• 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 $q\to\infty$? Commented May 22 at 20:52
• 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? Commented May 22 at 21:03
• 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! Commented May 22 at 21:35

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).

• Very clean way of writing this by taking the reciprocal from the beginning. Commented May 22 at 22:16

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 $$$$\tfrac{1}{\prod_{k=1}^n \tfrac{1}{1-p_k^{-q}}}\leq\tfrac{P_n}{Q_n}\leq \prod_{k=1}^n \tfrac{1}{1-p_k^{-q}}$$$$

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 $$q\to\infty$$.