TITLE: Asymptotics of the least common multiple of the first natural numbers
QUESTION [7 upvotes]: What is $$ \limsup_{n \to \infty} \frac{\log(\mathrm{lcm}(1,2, \dots, n))}{n} \ \ ?$$

REPLY [16 votes]: It is well-known that $\operatorname{lcm}(1,\ldots,n) = e^{\psi(n)}$, where $\psi$ is the Chebyshev's function. Since $\psi(x) = x + o(x)$, as $x \to +\infty$, (a form of the Prime Number Theorem) it follows that actually $\lim_{n \to +\infty} \frac{\log(\operatorname{lcm}(1,\ldots,n))}{n} = 1$. (See Part 1 of G. Tenenbaum - Introduction to Analytic and Probabilistic Number Theory).
Not really a MO question, in my opinion.

REPLY [3 votes]: Well, certainly $L=\text{lcm}(1,...,n)$ is divisible by $p^k\leq n$. We can take $k$ up to $[\log_p n]\geq \frac{\log n}{\log p}-1$. So $$\log L\geq \sum_{p\leq n}\left(\frac{\log n}{\log p}-1\right)\log p\approx n$$ by the Prime Number Theorem. So the lim sup should be 1.