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Me and a friend of mine worked on building a problem for AMM. It all started pretty well, but in the end we realized that the initial part of the solution was wrong. In few words, we thought we have proven that

$$\lim_{n \to \infty} \left(\log_{p_{n+1}} ((n+1)!)-\log_{p_n}(n!)\right)=1$$

where $p_n$ is the $n$-th prime number.

Denoting $x_n=\log_{p_n}(n!)$, there are a few ways that I think it is possible to prove that $x_{n+1}-x_n \to 1$:

First, maybe it is possible by direct computation (we tried and didn't get anything). The second method is by using Stolz-Cesàro in a different way: if we prove that $(x_{n+1}-x_n)$ is convergent then it should have the same limit as $x_n/n$ which converges to $1$.

So my question is:

Is it true that $(x_{n+1}-x_n) \to 1$, or at least $(x_{n+1}-x_n)$ is convergent?

Thank you.

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using the Prime number theory and the Stirling formula,I still can't decide whether this sequence is convergent or not. – user14242 Aug 12 '11 at 14:46
That's how we tried. :) I hope this is not an open question... – Beni Bogosel Aug 12 '11 at 14:49
I'd write $$\lim_{n \to \infty}(\log_{p_{n+1}}((n+1)!)-\log_{p_n}(n!))=1,$$ or even $$\lim_{n \to \infty}\Bigg(\log_{p_{n+1}}\Big((n+1)!\Big)-\log_{p_n}(n!)\Bigg)=1,$$ instead of $$\lim_{n \to \infty} \log_{p_{n+1}} ((n+1)!)-\log_{p_n}(n!)=1.$$ – Pierre-Yves Gaillard Aug 15 '11 at 12:24
OK. I'll change that. – Beni Bogosel Aug 15 '11 at 12:43
up vote 3 down vote accepted

I am going to assume that Harald Cramer's conjecture holds true, i.e. $g_n = p_{n+1}-p_n = \mathcal{O}(\log^2 p_n)$.

Let's further assume that the limit $\displaystyle\lim_{n\to \infty} \frac{g_n}{\log^2 p_n}$ exists and let $\displaystyle c = \lim_{n\to \infty} \frac{g_n}{\log^2 p_n}$ denote its value. Then $\displaystyle\lim_{n\to\infty} x_{n+1}-x_{n} = 1-c.$

To show this simplify the difference by using the factorial's recurrence equation: $$ x_{n+1}-x_n = \frac{\log p_n \cdot \log (n+1) - \log n! \cdot \log(1+\frac{g_n}{p_n})}{ \left(\log p_n + \log(1+\frac{g_n}{p_n}) \right) \log p_n } $$

Now use $\displaystyle\lim_{n\to \infty} \frac{\log n!}{p_n} = 1$. And write

$$ x_{n+1}-x_n = \frac{\log p_n \cdot \log (n+1) - \frac{\log n!}{p_n} \cdot p_n \log(1+\frac{g_n}{p_n})}{ \left(1 + \frac{1}{\log p_n} \log(1+\frac{g_n}{p_n}) \right) \log^2 p_n } $$ And, finally: $$ x_{n+1}-x_n = \frac{ \frac{\log (n+1)}{\log p_n} - \frac{\log n!}{p_n} \cdot \frac{g_n}{\log^2 p_n} \cdot \frac{p_n}{g_n} \log(1+\frac{g_n}{p_n})}{ \left(1 + \frac{1}{\log p_n} \log(1+\frac{g_n}{p_n}) \right) } $$

Now, since $\displaystyle\lim_{n\to\infty} \frac{g_n}{p_n} = 0$, $$\displaystyle\lim_{n\to\infty} \frac{p_n}{g_n} \log \left(1+\frac{g_n}{p_n}\right) = 1$$ and $$\displaystyle\lim_{n\to\infty} \left(1 + \frac{1}{\log p_n} \log \left(1+\frac{g_n}{p_n}\right) \right) = 0.$$

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If $g_n=O(\log^2 p_n)$ then it doesn't mean that the limit $g_n/(\log^2 p_n)$ exists. The ratio is only bounded. – Beni Bogosel Aug 12 '11 at 18:00
@Beni Thanks, then I guess I should add the assumption that the limit does exist. I will edit the post. – Sasha Aug 12 '11 at 18:03
Moreover, if there are infinitely many twin primes, then $g_n/\log^2 p_n$ has infimum $0$. So, it is likely that the lim inf is $0$. I have no intuition for whether the limit should exist, though. – David Speyer Aug 12 '11 at 18:25
On average, $g_n$ is $\log n$, so I think the infimum is zero without any unproved hypotheses. – Gerry Myerson Aug 12 '11 at 22:25
@Gerry Myerson Excellent point! – David Speyer Aug 15 '11 at 22:06

Let $m_{0}$ be the infimum of the set $\{m\in \mathbb{N} : \forall n \text{ sufficiently large, } g_{n}<(\log p_{n})^{m}\}$. Then one can prove than $m_{0}$ is not an integer. Thus it is sufficient to prove that $m_{0}<2$ to get $c=0$ and the desired limit.

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