$(n+1)!_\mathbb{P}$ and the Euler-Mascheroni constant 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.
 A: @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 
$$(n+1)!_\mathbb{P} \approx e^{n(C-\gamma)} n^n $$
