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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.

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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 $\{a_n: n \geq 1\}$.

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

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

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

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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

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

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

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