How to deduce a result assuming prime number theorem I am unable to derive this result assuming prime number theorem. Can someone please tell how to do it. 
Edit -> Here is a proof from stackexchange >but I couldn't think how last line is true. Can someone please  tell how it's true 

If $d_n = \text{lcm} \{1,2,3,... n\}$ then prove that the prime number theorem implies $$\lim _{n\to \infty} \frac {\log(d_n)} {n} =1$$


Can someone please tell how to derive it. 
 A: with
$lcm(1,2,3,...,x)= \prod\limits_{p\leq x} p^{max(\nu_p(1),\nu_p(2),\nu_p(3),...,\nu_p(x))}$
Since $$max(\nu_p(1),\nu_p(2),\nu_p(3),...,\nu_p(x))$$ is the maximum exponent of $p$ you'll find in numbers $\leq x$, it is $\lfloor \log_px \rfloor$
$$lcm(1,2,3,...,x)=\prod\limits_{p\leq x} p^{\lfloor \log_px \rfloor}$$
$$\log lcm(1,2,3,...,x)=\sum\limits_{p\leq x} {\lfloor \log_px \rfloor}\log p$$
and with
$\psi(x)=\sum\limits_{p\leq x}\lfloor \log_px \rfloor \log p $ (see EDIT)
$$\log lcm(1,2,3,...,x)=\log(d_x)=\psi(x)=\log e^{\psi(x)}$$
$$lcm(1,2,3,...,x)=e^{\psi(x)}$$
and from $\log(d_x)=\psi(x)$ you have your result

EDIT (not clear enought in the link):
$\psi(x)=\sum\limits_{p^k\leq x}\log p$
Since the maximum exponent of $p^k\leq x$ is $\lfloor \log_px \rfloor$ you have $\lfloor \log_px \rfloor$ powers of $p$ in that range: $\{p^1, p^2, p^3, ..., p^{\lfloor \log_px \rfloor}\}$, so $\log p$ is counted accordingly for each $p$:
$\psi(x)=\sum\limits_{p^k\leq x}\log p=\sum\limits_{p\leq x}\lfloor \log_px \rfloor \log p $
A: Well, I don't know why you do not mention books, if this troubles you. Here is an excerpt from Ingham. On average, if a mathematics book is old enough you can find a "preview" online, and this may include pages discussing your interest. 


