I'm studying the following limit

$$\lim_{n\to \infty} \frac{1}{n} \ln\left( \frac{(n+1)!_\mathbb{P}}{n^n}\right) $$

where $$(n+1)!_\mathbb{P} = \prod\limits_{p \in \mathbb{P}}^{} {p}^{\omega_p{(n)}},$$

where $\omega_p{(n)} = \sum_{k \geq 0} \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor$, and $\mathbb{P}$ is the set of primes. Here's what I have so far:

Before we take the limit, we can rewrite this as

$$\frac{1}{n} \ln\left( \frac{(n+1)!_\mathbb{P}}{n^n}\right) = \frac{\ln (n+1)!_\mathbb{P}}{n} - \ln(n) $$

$$= \frac{1}{n}\sum_{p \in \mathbb{P}}\sum_{k=0}^\infty \ln p \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor - \ln n$$

Using the property $ \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor\leq \left\lfloor x+y \right\rfloor$ (which I will assume holds over infinite sums, in spite of possible convergence issues), we have the inner sum such that

$$\sum_{k=0}^\infty \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor \leq \left\lfloor \frac{n}{p-1}\sum_{k=0}^\infty\frac{1}{p^k} \right\rfloor = \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor$$

Thus, we have

$$ \frac{1}{n}\sum_{p \in \mathbb{P}}\sum_{k=0}^\infty \ln p \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor - \ln n \leq \frac{1}{n}\sum_{p \in \mathbb{P}} \ln p \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor - \ln n$$

Now the next step I'm not so sure about; we assume we can do the following to the sum

$$\frac{1}{n}\sum_{p \in \mathbb{P}} \ln p \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor - \ln n $$

$$= \frac{1}{n}\left(\sum_{p \leq n} + \sum_{p > n } \right)\ln p \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor - \ln n $$

$$= \frac{1}{n}\sum_{p \leq n}\ln p \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor - \ln n + \frac{1}{n}\sum_{p > n } \ln p \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor$$

For simplicity sake, we denote $$ S_n = \frac{1}{n}\sum_{p > n } \ln p \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor$$

Now, using the identity $\left\lfloor x \right\rfloor \leq x$, we have

$$ = \frac{1}{n}\sum_{p \leq n}\ln p \left\lfloor \frac{n}{p-1} + \frac{n}{(p-1)^2} \right\rfloor - \ln n + S_n \leq \sum_{p \leq n} \frac{\ln p}{p-1} + \frac{\ln p}{(p-1)^2} - \ln n + S_n $$

Now we take the limit; using the following Euler-Mascheroni constant identity $$\gamma = \lim_{n \to \infty} \left( \ln n - \sum_{p \le n} \frac{ \ln p }{ p-1 } \right)$$

We have

$$\lim_{n\to \infty} \frac{1}{n} \ln\left( \frac{(n+1)!_\mathbb{P}}{n^n}\right) \leq \lim_{n \to \infty} \sum_{p \leq n} \frac{\ln p}{p-1} + \frac{\ln p}{(p-1)^2} - \ln n + S_n $$

$$ = C-\gamma + \lim_{n \to \infty} S_n$$

where $C = \sum_{p \in \mathbb{P}} \frac{\ln p}{(p-1)^2}$. Assuming $ \lim_{n \to \infty} S_n = 0$, we have that for large $n$,

$$(n+1)!_\mathbb{P} \approx e^{n(C-\gamma)}n^n$$

which is pretty neat. I have a few questions now

$$ \text{1) How can we prove/disprove} \lim_{n \to \infty} S_n = 0 \text{?}$$ $$ \text{2) Is the step that I'm insecure about allowed?}$$

Thank you for your patience, I really appreciate the support.

  • 1
    $\begingroup$ It looks like your $S_n$ has at most 1 non-zero summand. $\endgroup$ Dec 17, 2015 at 6:41

1 Answer 1


@Fedor Petrov, I think you're right. I also tweaked my approach a bit to tighten $S_n$.

Let $$I_n = \frac{1}{n} \ln\left(\frac{(n+1)!_\mathbb{P}}{n^n}\right)$$ Of course, we want to estimate $\lim_{n \to \infty} I_n $. Let $$f_n(p) = \ln p\sum_{k=0}^\infty \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor$$

Repeating the above, and redefining $S_n$, we have

\begin{align} I_n & = \frac{\ln (n+1)!_\mathbb{P}}{n} - \ln n \\ & = \frac{1}{n} \sum_{p \in \mathbb{P}} f_p(n) -\ln n \\ & = \frac{1}{n} \sum_{p \leq n} f_p(n) - \ln n + \overbrace{\frac{1}{n} \sum_{p > n} f_p(n)}^{= S_n}\\ & \leq \sum_{p \leq n} \frac{\ln p}{(p-1)^2} - \left(\ln n - \sum_{p \leq n} \frac{\ln p}{p-1} \right)+ S_n \end{align}

Now, notice

$$S_n = \frac{1}{n} \sum_{k\geq 0} \sum_{p>n} \ln p \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor$$

The infinite sum essentially collapses, since, for $k > 0$, $p>n$ and $p$, $n$ are positive terms, so $\left\lfloor \frac{n}{(p-1)p^k}\right\rfloor = 0$. Thus, the sum reduces to

$$S_n = \frac{1}{n} \sum_{p>n} \ln p \left\lfloor \frac{n}{p-1} \right\rfloor$$

Here, we only need to consider the smallest prime such that $p>n$; we essentially have the following for the sum

\begin{align} S_n = \begin{cases} 0, & p-1>n \\ \frac{\ln p}{n}, & p-1=n \\ \end{cases} \end{align}

The smallest prime such that $p>n$ implies that $p_{\pi(n)+1} > n$, where $\pi(n)$ is the prime counting function, and $p_n$ is the $n$-th prime; this gives us something explicit to work with. Using the fact $\pi(n)$ is the right inverse of $p_n$, we have then

\begin{align} S_n = \begin{cases} 0, & \pi(n)+1>\pi(n+1) \\ \frac{\ln p_{\pi(n)+1}}{n}, & \pi(n)+1=\pi(n+1) \\ \end{cases} \end{align}

Now consider whenever $\pi(n)+1=\pi(n+1)$ (don't how often this happens, but surely infinitely often?); practically we want to look at

$$\lim_{n \to \infty} \frac{\ln p_{\pi(n)+1}}{n} $$

Using the result, $n >5$, $$ n h_{-1}(n) < p_n < n h_{0}(n)$$ where $h_a(n) = \ln(n)+\ln(\ln(n)+a)$ grows logarithmically, and using the prime number theorem, we have

$$ \frac{1}{n}\left(\ln\left(\frac{n}{\ln(n)}+1\right) +\ln h_{-1}(n) \right) < \frac{\ln p_{\pi(n)+1}}{n} < \frac{1}{n} \left(\ln\left(\frac{n}{\ln(n)}+1\right) +\ln h_{0}(n) \right)$$

Letting $n \to \infty$, and applying the squeeze theorem, we have then

$$\lim_{n \to \infty} \frac{\ln p_{\pi(n)+1}}{n} = 0 \implies \lim_{n \to \infty} S_n = 0$$

which gives us our desired result that

$$(n+1)!_\mathbb{P} \approx e^{n(C-\gamma)} n^n $$

  • 1
    $\begingroup$ Isn't $\pi(n)+1 = \pi(n+1)$ exactly when $n+1$ is prime, which certainly happens infinitely often?! $\endgroup$ Jan 18, 2016 at 5:27
  • $\begingroup$ @GeoffRobinson How could I have missed that?! Thank you for pointing that out! $\endgroup$
    – Brian
    Jan 18, 2016 at 18:44

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