I need to prove there are constants $c_1,c_2 > 1$ such that $$ c_1^n < \mathrm{lcm}(1,\dots,n) < c_2^n $$ for $n$ integer, $n \geq 2$.
I tried to use that $\mathrm{lcm}(1,\dots,n)=\prod_{p \leq n} p ^{b_p}$ where $b_p$ is the greatest integer such that $p^{b_p} \leq n$. Maybe $c_2 = 2$? No ideas about the value of $c_1$ though. Any clue would be appreciated.
Note that $\log\text{lcm}(1,2,\ldots,n)=\psi(n)$ (the second Chebyshev function). You want to show that $\log c_1\leqslant\frac{\psi(n)}n\leqslant\log c_2$ for some $c_1,c_2>1$, that is, $\log c_1,\log c_2>0$.
We don't need the full strength of the PNT (which is equivalent to $\psi(n)\sim n$) to show this, in fact this is one of Chebyshev's estimates:
Chebyshev proved it (and the proof is not that hard, see for example Hildebrand's notes, Theorem 3.1 page 83) for $\log c_1=\frac12$ and $\log c_2=2+\epsilon$, $\epsilon>0$. A very beautiful alternative proof for the lower bound with $c_1=2$ is given in $\text{lcm}(1,2,3,\ldots,n)\geq 2^n$ for $n\geq 7$.
Better bounds are implied by the PNT (and RH, as you wish). PNT gives $c_1=e-\epsilon$ and $c_2=e+\epsilon$, $\epsilon>0$. In fact, $\text{lcm}(1,2,\ldots,n)\sim e^n$.
Actually, one can show that the prime number theorem implies that the limit $$\lim_{n \rightarrow \infty}\sqrt[n]{\text{lcm}(1,2,\dots,n)} $$ exists, so this would solve the problem. So assuming the PNT, this problem is not too hard. :)
