Series convergence by Gauss Test I want to try to prove that the $$\sum_{n=1}^\infty \frac{n!e^n}{n^{n+p}}$$  series convergent for $p>1.5$ with Gauss Test but failed? Gauss Test said if
$\frac{a_n}{a_{n+1}}$ can be represnted as
$\frac{a_n}{a_{n+1}}=\lambda+\frac{\mu}{n}+b_n$
where $\sum_{n=1}^\infty b_n$ absolutely convergent
then

*

*$\lambda>1$ convergent

*$\lambda<1$ divergent

*$\lambda=1, \mu>1 $ convergent

*$\lambda=1, \mu\leq1 $ divergent

My work: I found that
$$\frac{a_n}{a_{n+1}}=\frac{1}{e}{(1+\frac{1}{n})}^{n+p}=\frac{1}{e}(1+1/n)^{n+p}=\frac{1}{e}(1+\frac{n+p}{n}+\frac{(n+p)(n+p-1)}{2}\frac{1}{n^2}+\dots)),$$ and then stopped and didnt see a continuation.
 A: Note that, as $n$ goes to infinity,
$$\frac{a_n}{a_{n+1}}=\frac{n!e^n}{n^{n+p}}\cdot \frac{(n+1)^{n+1+p}}{(n+1)!e^{n+1}}=\frac{(1+\frac{1}{n})^{n+p}}{e}=1+\frac{p-\frac{1}{2}}{n}+O(1/n^2)$$
where
\begin{align}
\left(1+\frac{1}{n}\right)^{n+p}&=\exp\left((n+p)\ln(1+\frac{1}{n})\right)=\exp\left((n+p)(\frac{1}{n}-\frac{1}{2n^2}+O(1/n^3))\right)\\&=\exp\left(1+\frac{p-\frac{1}{2}}{n}+O(1/n^2)\right)
=e\left(1+\frac{p-\frac{1}{2}}{n}+O(1/n^2)\right).\end{align}
Hence, according to Gauss Test, the series is convergent when $p-\frac{1}{2}>1$, that is $p>\frac{3}{2}$.
A: Proof with Stirling formula :
You have
$$\frac{n!e^n}{n^{n+p}} \sim \sqrt{2\pi n} \frac{n^ne^n}{e^nn^{n+p}} \sim \frac{\sqrt{2\pi}}{n^{p-\frac{1}{2}}}$$
which is the general term of a convergent series, since $p>1,5$. So the given series converges.
Proof with Gauss test :
Let $$a_n = \frac{n!e^n}{n^{n+p}}$$
You have
$$\frac{a_n}{a_{n+1}} = \frac{n!e^n(n+1)^{n+1+p}}{(n+1)!e^{n+1}n^{n+p}} = \frac{1}{e}\left(\frac{n+1}{n} \right)^{n+p} = \exp \left[(n+p)\ln \left( 1+\frac{1}{n}\right)-1 \right]$$
i.e.
$$\frac{a_n}{a_{n+1}} =  \exp \left[(n+p)\left(\frac{1}{n} -\frac{1}{2n^2} + O\left( \frac{1}{n^3}\right) \right)-1 \right] =  \exp \left[ \frac{p-\frac{1}{2}}{n} + O\left( \frac{1}{n^2} \right)  \right]$$ i.e. $$\frac{a_n}{a_{n+1}}= 1+\frac{p-\frac{1}{2}}{n}+O\left(\frac{1}{n^2}\right)$$
Now you can conclude using Gauss test.
