I want to show that \begin{eqnarray} \frac{n^n}{n!e^n} \end{eqnarray} whether it is decreasing or increasing sequences or neither.
$\lim_{n \to \infty} \frac{u_{n+1}}{u_n}=\lim_{n \to \infty}(1+\frac{1}{n})^n\frac{1}{e}$.
Since $\lim_{n \to \infty}(1+\frac{1}{n})^n=e$, $\lim_{n \to \infty} \frac{u_{n+1}}{u_n}=1$
What can we say about the sequences according to $\lim_{n \to \infty} \frac{u_{n+1}}{u_n}=1$.
Can we say it is neither decreasing nor increasing ? Thank you for your help.
Use the fact that $(1+\frac 1 n)^n$ is increasing and $\lim (1+\frac1n)^n = e$ therefore $(1+\frac1n)^n \lt e$ and $\frac{u_{n+1}}{u_n} \lt 1$
The sequence is non-increasing. Before proving it, first: compute the first terms, it will help you check whether your intuition is correct!
On Mathematica or Wolfram:
DiscretePlot[ n^n/(n! E^n), {n, 1, 100}][
]1
Then, once you have done that and gotten a sense of what you have to prove (here: the sequence, if anything, does look decreasing) you can attempt to prove it.
Let $a_n \stackrel{\rm def}{=} \frac{n^n}{n!e^n}$. Then it is easy to see that $$ \frac{a_n}{a_{n+1}} = e\left(\frac{n}{n+1}\right)^n = \frac{e}{(1+\frac{1}{n})^n} $$ Now, from $$ (1+\frac{1}{n})^n = e^{n\ln(1+\frac{1}{n})} \leq e^{n\cdot \frac{1}{n}} =e^1 =e $$ (which follows from the standard inequality $\ln(1+x)\leq x$) we get $$ \frac{a_n}{a_{n+1}} \geq 1 $$ for all $n\geq 1$.
If $a(n) = n^n/(n! e^n)$, then
$$ \eqalign{\ln \left( \frac{a(n+1)}{a(n)} \right) &= n \log(1+1/n) - 1\cr &= -\frac{1}{2n} + \frac{1}{3n^2} - \frac{1}{4n^3} + \ldots}$$ I claim this is negative. Indeed, for $n \ge 1$, $$-\frac{1}{2m n^{2m}} + \frac{1}{(2m+1)n^{2m+1}} < 0 $$ Thus $a(n+1)/a(n) < 1$, and the sequence is decreasing.
If $n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$, then:
$$\frac{n^n}{n!e^n}=\frac{1}{n!} \cdot \left(\frac{n}{e}\right)^n\sim \left(\frac{e}{n}\right)^n \frac{1}{\sqrt{2\pi n}} \cdot \left(\frac{n}{e}\right)^n = \frac{1}{\sqrt{2\pi n}}$$ which is decreasing.
