How to prove that the series $\sum\ n^s e^{-n}\ \ (s\ge 0)$ converges? 
Prove that $\ \sum n^s \cdot e^{-n}  , \ s \ge 0$ converge.

My attempt using ratio test:
$$\ \frac{A_{n+1}}{A_n} = \frac{(n+1)^s}{e^{n+1}} \cdot \frac{e^n}{n^s} = \frac{(n+1)^s}{e\cdot n^s} $$
But how can I proceed from here?
 A: Rewrite $$\frac{(n+1)^s}{e \cdot n^s}$$ as $$\left(\frac{n+1}{n}\right)^s \cdot \frac{1}{e}$$
If you need the next step, divide the top and bottom of the fraction by n to get
$$\left(1+\frac{1}{n}\right)^s \cdot \frac{1}{e}$$
A: Hint The root test is more obvious here: 
$$\lim_n \sqrt[n]{n^s \cdot e^{-n}}=  \lim_n (\sqrt[n]{n})^s e^{-1}= 1^s \cdot e^{-1}=\frac{1}{e}$$
A: Since $n^{1/n} \to 1$,
for fixed $s$,
$n^{s/n} \to 1$
so
$n^{s/n} < 2$
for all large enough $n$.
Therefore
$\begin{array}\\
n^se^{-n}
&=(n^{s/n})^{n}e^{-n}\\
&=(\frac{n^{s/n}}{e})^n\\
&<(\frac{2}{e})^n
\qquad\text{for all large enough } n\\
\end{array}
$
and the sum of these converges.
A: We can also use the Cauchy's convergence test. This test tells us that a serie $\sum_{k=1}^\infty a_k$ is convergent if, and only if, for all $\epsilon>0$ there is a $N_\epsilon\in\mathbb{N}-\{0\}$ such that
$$
n> N_\epsilon \implies | a_{k+1}+a_{k+2}+\ldots+a_{k+p}|<\epsilon \quad \forall p\in\mathbb{N}
$$
Note that
\begin{align}
\left|
(n+1)^s e^{-(n+1)}+\cdots+(n+p)^s e^{-(n+p)}
\right|
=&
\sum_{\ell=1}^{p}(n+\ell)^s e^{-(n+\ell)}
\\
\leq&
\sum_{\ell=1}^{p}(n+p)^s e^{-(n+\ell)}
\\
\leq&
\sum_{\ell=1}^{p}(n+p)^s e^{-n}
\\
=&
p(n+p)^s e^{-n}
\\
=&
\frac{n^s}{e^n}
\cdot
\underbrace{
p(1+p/n)^s
}_{\mathrm{limited}}
\end{align}
Choose $ N_{p,s}\in\mathbb{N}$  big enough so that $n> N_{p,s}$ implies $(1+p/n)^s<2$. Then $n>N_{p,s}$ implies
\begin{align}
\left|
(n+1)^s e^{-(n+1)}+\cdots+(n+p)^s e^{-(n+p)}
\right|
\leq &
\frac{2pn^s}{e^n}
\end{align}
For an arbitrary $ \epsilon> 0 $ we can choose a large enough $ N_{p,s,\epsilon} $ such that $n>N_{p,s,\epsilon}$ implies $$\frac{2pn^s}{e^n}<\epsilon.$$  Choose $N_{\epsilon}=\max\{N_{p,s,\epsilon},N_{s,p}\}$. Therefore, the convergence of the series follows .
