TITLE: why $ \rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$
QUESTION [15 upvotes]: Let $n\ge 7$ be positive integers,show that
$$f(n)=\rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$$
Anyone know this problem background?or maybe have best proof or best result?

REPLY [7 votes]: the $2^n$ side is answered at $\text{lcm}(1,2,3,\ldots,n)\geq 2^n$ for $n\geq 7$ 
The governing phenomenon here is that the LCM in question is $e^{\psi(n)},$ where $\psi(n)$ is Chebyshev's second function. So, from the text comments for https://oeis.org/A003418 , we find

The prime number theorem implies that LCM(1,2,...,n) = exp(n(1+o(1)))
  as n -> infinity. In other words, log(LCM(1,2,...,n))/n -> 1 as n ->
  infinity. - Jonathan Sondow, Jan 17 2005

We can be very specific with the upper bound, Rosser and Schoenfeld (1962) Theorem 12 on page 71,
$$ \psi(x) < 1.03883 \, x  $$ for $x > 0.$ As $e^{1.03883} \approx 2.8259087,$ we find
$$ \operatorname{lcm} (1,2,3,\ldots,n) < 2.826^n $$
This bound is tightest at $n=113,$ which is where the maximum of $\psi(x)/x$ is obtained.
Apparently, Hanson (1972) showed that $ \operatorname{lcm} (1,2,3,\ldots,n) < 3^n, $ see references in http://arxiv.org/abs/0906.2295