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?

Unless I'm misinterpreting something here, $1$ only has itself as a divisor, hence $\gcd(1,\dots)=1$ regardless. — Simply Beautiful Art, Mar 11 '20 at 18:36
Did you mean LCM? If so, note each prime $p\le n$ contributes a prime power factor $\le n$. By the PNT, there are asymptotically $n/\ln n=\log_ne^n$ of these. — J.G., Mar 11 '20 at 18:36

Gary · Accepted Answer · 2025-02-07T08:27:42.367

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

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

1 Answers1

Linked