TITLE: least common multiple of $\{1,2,...,n\}$ is bigger than $2^{n-1}$
QUESTION [13 upvotes]: The least common multiple of $\{1,2,...,n\}$ is greater than $2^{n-1}$ for any $n \ge 3$.

I found this in a MATHEMATICA book, but I don't know how to prove this. Can you help me?
[Edit: This thread has a discussion of an asymptotic stronger result, but that relies on the Prime Number Theorem. What else is known about this? JL]

REPLY [15 votes]: Here is a neat proof:
Consider the product
$$\text{lcm}(1, 2, \dots, 2n + 1)\int_0^1 (x)^n(1 - x)^n \text{d}x$$
This is an integer, because expanding using the binomial theorem and integrating each part gives an integer.  Additionally, the integral is less than
$$\int_0^1 \left(\frac14\right)^n \text{d}x = \frac{1}{4^n}$$
so
$$\text{lcm}(1, 2, \dots, 2n + 1) > 4^n = 2^{2n}$$
which proves the desired inequality for odd values.
To prove it for even values, consider the product
$$\text{lcm}(1, 2, \dots, 2n)\int_0^1 (x)^{n}(1-x)^{n-1} \text{d}x$$
This product is also an integer.  Additionally, the integral is less than
$$\int_0^1 x\left(\frac14\right)^{n-1}\text{d}x = \frac{1}{2\cdot4^{n-1}}$$
so we arrive at
$$\text{lcm}(1, 2, \dots, 2n) > 2\cdot4^{n-1} = 2^{2n - 1}$$
which proves the desired result for even values.