# Prove there are two positive constants for inequality

Prove, that there are two positive constants $$C_{1}$$ and $$C_{2}$$ such that $$C_{1}\sqrt{n}\left(\frac{n}{e}\right)^{n} < n! < C_{2}\sqrt{n}\left(\frac{n}{e}\right)^{n}$$ So I know there is Stirling's approximation $$n! \sim \sqrt{2\pi}\sqrt{n}\left(\frac{n}{e}\right)^{n}$$. Also I was given an indication, that it may be better to take a logarithm of $$n!$$ and estimate the error of approximation for $$\int_{1}^{n} lnx \space dx$$ by using Trapezoidal rule for $$1 < 2 < \dots < n$$ partition. But I don't understand how to use all of that to prove what I need.

This is obvious from Stirling's approximation. Let $$a_n=\frac {n!} {\sqrt n (\frac n e)^{n}}$$. Then $$a_n \to \sqrt {2\pi}$$ and hence the inequality holds for suitable $$C_1$$ and $$C_2$$. You can take $$C_1$$ and $$C_2$$ to be the infimum and the supremum of $$\{a_n: n \geq 1\}$$.

$$\log ({n!}) - \log(\sqrt{2\pi}\sqrt{n}(\frac{n}{e})^n)$$

= $$\sum_{k=1}^n \log k - \log\sqrt{2\pi} - \frac{1}{2}\log(n) - n \log(\frac{n}{e})$$.

Now you may approximate the error by estimating $$\sum_{k=1}^n \log k$$ by $$\int_1^n \log x dx$$.

You should take the measure of the error in stirling approximation and espress it in terms

$$n! = \sqrt{2\pi}\sqrt{n} (\frac{n}{e})^n + \epsilon_1 = (\sqrt{2\pi} + \epsilon_2 )\sqrt{n} (\frac{n}{e})^n$$ and with $$\epsilon_1$$ decreasing with n increase (just take max $$\epsilon_1$$) later set

$$C_1 < (\sqrt{2\pi} + \epsilon_2 ) < C_2$$

• Please set a good mark for a good answer – user135617 Apr 10 at 14:45