Is $\displaystyle \lim_{n \rightarrow \infty} \frac{[1,2, \cdots, n]}{3^n} = 0$? 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 $c$?
 A: The Prime Number Theorem says that, to the extent that it makes sense, your LCM is about $$  e^n. $$
In fact, all that has been proved is that  $$ \lim_{n \rightarrow \infty} \; \; \; \frac{\log (\operatorname{lcm}(1,2,3,...,n))}{n} = 1 $$
Indeed, we have Chebyshev's second function $\psi $ and
$$ \psi(n) = \log (\operatorname{lcm}(1,2,3,...,n)) $$
Hardy and Wright show that the Prime Number Theorem is equivalent to the assertion that 
$$  \psi(x) \sim x,  $$ meaning
$$  \lim_{x \rightarrow \infty}  \frac{\psi(x)}{x} = 1. $$
They then prove this as their Theorem 434.
Combining a few results from Rosser and Schoenfeld (1962), for $x \geq 41,$
$$    1 - \frac{1}{\log x} \; \; \;  \; < \; \frac{\psi(x)}{x} \; < \; \; \; \;  1 + \frac{1}{2 \log x} + \frac{1.42620}{\sqrt x}$$
I ought to emphasize  that we know very little about
$$  \frac{e^{\psi(x)}}{e^x},  $$
which is sometimes enormous and sometimes tiny.
A: 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, $\lim_{n\rightarrow\infty}\frac{\text{LCM}\{1,2,...,n\}}{3^n}=0$.
