TITLE: Is $\displaystyle \lim_{n \rightarrow \infty} \frac{[1,2, \cdots, n]}{3^n} = 0$?
QUESTION [5 upvotes]: I'm trying to study the increasing speed of the l.c.m of $\{1,2,\cdots,n\}$.
It's easy to see that it increases more rapidly than $n$ and more slowly than $n!$. But how about $c^n$ for a positive constant $c$?

REPLY [3 votes]: We can write $\text{LCM}\{1,2,...,n\}=\prod_{p\in\text{primes}}\ p^{\lfloor \log_p n\rfloor}$. If you need some intuition behind this, take each prime and take the largest power $\le n$, and see why the product of these primes up to $n$ is exactly the $\text{LCM}$ of $1,2,...,n$.
Note that $\text{LCM}\{1,2,...,n\}=\prod_{p\in\text{primes}}\ p^{\lfloor \log_p n\rfloor}\le
\prod_{p\in\text{primes}}
n \thicksim n^{n/\ln n}$ by Prime Number Theorem.
Following,
$$\lim_{n\rightarrow\infty}\frac{\text{LCM}\{1,2,...,n\}}{3^n}\le \lim_{n\rightarrow\infty}\frac{n^{n/\ln n}}{3^n}=0$$
Since each term is positive, by the squeeze theorem, $\lim_{n\rightarrow\infty}\frac{\text{LCM}\{1,2,...,n\}}{3^n}=0$.