<s>[INST] Existence of limit $\lim_{n\to\infty}\frac{n!}{n^{n+1/2}e^{-n}}$ It is shown that
$$
 e^{\frac{11}{12}}n^{n+1/2}e^{-n}< n!<en^{n+1/2}e^{-n}.
$$
The question is, how to conclude from here that the limit
$$
\lim_{n\to\infty}\frac{n!}{n^{n+1/2}e^{-n}}
$$
exists? I suspect that $\frac{n!}{n^{n+1/2}e^{-n}}$ is decreasing, but I can't prove this claim.
Note: I am studying a topic about Stirling's formula, so please avoid mentioning that the limit exists due to that formula.
 [/INST] You can prove the monotonicity as follows. Let $a_n  = \frac{{n!}}{{n^{n + 1/2} e^{ - n} }}$. Then
$$
\frac{{a_n }}{{a_{n + 1} }} = \left( {1 + \frac{1}{n}} \right)^{n + 1/2} \frac{1}{e} = \exp \left( {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right).
$$
But
\begin{align*}
\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1 &= (2n + 1)\frac{1}{2}\log \left( {\frac{{1 + \frac{1}{{2n + 1}}}}{{1 - \frac{1}{{2n + 1}}}}} \right) - 1 \\ & = (2n + 1)\tanh ^{ - 1} \left( {\frac{1}{{2n + 1}}} \right) - 1,
\end{align*}
and $\tanh ^{ - 1} (x) = x + \frac{{x^3 }}{3} + \frac{{x^5 }}{5} +  \ldots  > x$ for $0<x<1$, i.e.,
$$
\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1 > (2n + 1)\frac{1}{{2n + 1}} - 1 = 0.
$$
Consequently,
$$
\frac{{a_n }}{{a_{n + 1} }} > e^0  = 1,
$$
i.e., the sequence $a_n$ is strictly decreasing.
</s>