Proof that lcm$(1, \ldots, n)=O(e^n)$

Question

I'm trying to prove that lcm$$(1, \ldots, n)=O(e^n)$$

I know that lcm$(1, \ldots, n)^{1/n}$ converges to $e$ so given any $\varepsilon >0$ there exists some positieve integer $n_0$ such that if $n \geq n_0$ then $$\mathrm{lcm}(1, \ldots, n) \leq (e+\varepsilon)^n$$ but I'm not able to get the inequality just for $e$. How can you get such inequality?

Answer 1

I assume, you mean $\mathrm{lcm}$ instead of $\mathrm{gcd}$. In terms of the Chebyshev function $$ {\mathop{\rm lcm}\nolimits} (1,2, \ldots ,n) = e^{\psi (n)} . $$ By a theorem of E. Schmidt, there is a positive constant $K$, such that for infinitely many natural numbers $n$, $$ e^{\psi (n)} > e^{n + K\sqrt n } . $$ Hence, ${\mathop{\rm lcm}\nolimits} (1,2, \ldots ,n)=\mathcal{O}(e^n)$ cannot hold for all sufficiently large $n$.