1. ## limit

Calculate this limit ( n ــــ > + infinity ) :

$
\frac{n!e^{n}}{n^{n+\frac{1}{2 }}}
$

2. Hello, dhiab!

$\lim_{x\to\infty} \frac{n!e^{n}}{n^{n+\frac{1}{2 }}}
$

Ratio Test: . $\left|\frac{a_{n+1}}{a_n}\right| \;\;=\;\;\left|\frac{(n+1)!e^{n+1}}{n^{n+\frac{3}{ 2}}} \cdot \frac{n^{n+\frac{1}{2}}}{n!e^n}\right| \;\;=\;\left|\frac{(n+1)!}{n!}\cdot \frac{e^{n+1}}{e^n} \cdot\frac{n^{n+\frac{1}{2}}}{n^{n+\frac{3}{2}}} \right|$

. . . . . . . . $=\;\; \left|\frac{n+1}{1}\cdot\frac{e}{1} \cdot\frac{1}{n}\right| \;\;=\;\;\left|\frac{n+1}{n}\cdot e\right|$

Divide top and bottom by $n\!:\quad \left|\frac{1 + \frac{1}{n}}{1}\cdot e\right|$

Then: . $R \;=\;\lim_{x\to\infty}\left|\frac{1 + \frac{1}{n}}{1}\cdot e\right| \;=\; \left|\frac{1+0}{1}\cdot e\right| \;=\;e$

Since $R > 1$, the series diverges.

3. ## Possible error

Soroban,

I think you made an error in your ratio test: since $a_n=\frac{n!e^n}{n^{n+\frac{1}{2}}}$,
then
$a_{n+1}=\frac{(n+1)!e^{n+1}}{\left(\color{red}n+1\ color{black}\right)^{n+\frac{3}{2}}}$,
which prevents the cancelling you did.

Dhiab, the key to this limit is Stirling's approximation:
$n!\approx\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}$ for large n.

--Kevin C.