Does $\log(p \#)\approx p$ hold for large $p$?

Question

This program

? for(k=1,9,s=0;forprime(j=1,10^k,s=s+log(j));print(k,"  ",s))
1  5.347107530717468680518589435
2  83.72839039906392294502692284
3  956.2452651200588678124015162
4  9895.991379156987312668949617
5  99685.38926861255083662384712
6  998484.1750256342921339727141
7  9995179.317856311896844315932
8  99987730.01802200438320982241
9  999968978.5775661447989107185  

calculates the natural logarithms of the primorials $10 \#,100 \#,...,10^9 \#$

The numerical analysis leads to the conjecture $\log(p \#)\approx p$ for large $p$.

• Is it true that $$\lim_{p\rightarrow \infty} \frac{\log(p \#)}{p}=1?$$

• If yes, can this result be derived from the prime number theorem?

• Is there a series giving very good approximations of $\log(p \#)$ for large $p$, for example $\log(10^{100} \#)$ ?

Answer 1

One has $\log (p\#)$ is the sum of $\log q$ over all primes up to $p$. This is basically the (first) Chebyshev function $\theta(x)$.

It is a well-known reformulation of the Prime Number Theorem that $\theta(x) \sim x$ that is $\theta(x)/ x$ tends to $1$. Indeed, frequently a proof of the prime number theorem actually derives this and from it derives the asymptotic for the prime counting function.