How can we formally / rigorously use Mertens' third theorem with $n^2 - 1$ instead of $\ln n$?

Question

I'm quite new to Analytic NT, so was wondering if the following is true, and beyond being obviously true, how could we prove it rigorously line-by-line?

See: Mertens Third theorem

The function they use to "modulate the product" (which normally diverges to zero) is $\ln n$. And the constant they get is $\gt 1/2$. Then to me, it's obviously true that if we used the function $n^2 -1$ in place of $\ln n$ that either:

• The constant approached in the limit is easily $\gt 1/2$.
• Or, the limit approached is $\infty$.

But I have no idea how to "work with asymptotics". This seems like it would be an easy exercise. Do we even need a proof for it?

I'd really like to know how to work with the limits involved.

Answer 1

\begin{align} \lim_{n\to\infty}(n^2-1)\prod_{p\le n}\left(1-\frac{1}{p}\right) &= \lim_{n\to\infty}\frac{n^2-1}{\log n}\log n\prod_{p\le n}\left(1-\frac{1}{p}\right)\\ &=\lim_{n\to\infty}\frac{n^2-1}{\log n}\cdot\lim_{n\to\infty}\log n\prod_{p\le n}\left(1-\frac{1}{p}\right)\\ &=e^{-\gamma}\lim_{n\to\infty}\frac{n^2-1}{\log n} = \infty \end{align}

From this proof is easy to see that if you change $\log n$ by any function that grows even slightly faster than $\log n$ (for example $\log n\cdot \log\log n$) the limit will be infinite.

Answer 2

In fact, the number theory here is completely irrelevant. It’s “just” an analysis question. As you suspected, it is not difficult; I cannot urge you enough to study limits inside out (any analysis text should provide you with enough material) before moving on towards analytic number theory.

Anyway, let $u_n$ be a sequence of positive real numbers and suppose that $u_n \ln{n} \rightarrow c$ for some $c > 0$ (here, of course, $c=e^{-\gamma}$, and $u_n=\prod_{p \leq n}{\left(1-\frac{1}{p}\right)}$).

This means that there exists $N_0 > 0$ such that if $n \geq N_0$, then $u_n \ln{n} \geq c/2$.

Now, for all $n >1$, $(n^2-1)u_n=\frac{n^2-1}{\ln{n}}\times u_n \ln{n}$.

So, for all $n >N_0+1$, $(n^2-1)u_n \geq \frac{n^2-1}{\ln{n}}\times c/2$.

It is a fact that $\ln{n} \leq n-1$ for all integers $n \geq 1$. You can prove it by concavity (if you know it), or by taking the logarithm of $e^{n-1} = 1+(n-1)+\frac{(n-1)^2}{2}+ \ldots \geq 1+(n-1)=n$).

Hence, for all $n > N_0+1$, $(n^2-1)u_n \geq (n+1)c/2$. In particular, $(n^2-1)u_n \rightarrow \infty$.

---

There is a more direct argument in the specific situation that you are considering:

$$(n^2-1)\prod_{p \leq n}{(1-\frac{1}{p})} \geq (n^2-1)\prod_{2 \leq k \leq n}{(1-\frac{1}{k})} \geq (n^2-1)\prod_{2 \leq k \leq n}{\frac{k-1}{k}} = \frac{n^2-1}{n} \geq \frac{n^2-1}{n+1}=n-1,$$

whence the limit as $n \rightarrow \infty$ is $+\infty$.