Convergence of $\sum_{n=1}^\infty\frac {n^{n}}{e^nn!}$ 
Check the convergence of: $\displaystyle\sum_{n=1}^\infty\frac {n^{n}}{e^nn!}$

Using the root test I get: $\displaystyle\lim_{n \to\infty} \dfrac {n}{e\sqrt[n]{n!}}$ now I'm left with showing that $n > \sqrt[n]{n!} \ \ \forall n$, can I just raise it to the power of $n$ like so:  $\ n^n>n!$ ?
Alternatively, using limit arithmetic: $\displaystyle\lim_{n \to\infty} \dfrac {n}{e\sqrt[n]{n!}}=\displaystyle\lim_{n \to\infty} \dfrac {1}{\large\frac e n \sqrt[n]{\frac {n!}{n^n}}}>1$ (that's not very persuasive I know) so it diverges. 
Edit: Root test won't work.
Note: Stirling, Taylor or integration are not allowed. 
 A: Let
$$u_n=\frac{n^n}{e^nn!}$$
so
$$\frac{u_{n+1}}{u_n}=\frac1e \left(1+\frac1n\right)^n=\frac1e\exp\left(n\left(\frac1n-\frac1{2n^2}+o\left(\frac1{n^2}\right)\right)\right)=\exp\left(-\frac1{2n}+o\left(\frac1{n}\right)\right)\\=1-\frac1{2n}+o\left(\frac1{n}\right)$$
so using the Raabe-Duhamel's rule the series is divergent.
A: An idea: put
$$a_n:=\frac{n!}{n^n}\implies \frac{a_{n+1}}{a_n}=\frac{(n+1)!n^n}{(n+1)^{n+1}n!}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1$$
Thus, the series $\;\sum a_n\;$ converges by the quotient text, and from here
$$a_n=\frac{n!}{n^n}\xrightarrow[n\to\infty]{}0$$
and from here, for some $\;n\;$ on, we get that
$$n!\le n^n\implies\sqrt[n]{n!}\le n$$
A: Let us give another proof:
Lemma: $\sum a_n$, $\sum b_n$ are series of positive terms. If we have $\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n}$ then if $\sum b_n$ is convergent so is $\sum a_n$.
proof: $\frac{a_{n+1}}{a_n}\leq \frac{b_{n+1}}{b_n} \Rightarrow \frac{a_{n+1}}{b_{n+1}}\leq \frac{a_n}{b_n}$.
Consider $c_n=\frac{a_n}{b_n}$ then $\frac{a_{n+1}}{b_{n+1}}\leq \frac{a_n}{b_n}\Rightarrow c_{n+1}\leq c_n\leq c_1\Rightarrow \frac{a_n}{b_n}\leq c_1 \Rightarrow a_n \leq c_1 b_n$.
Now $\sum b_n$ is convergent so is $\sum a_n$[by comparison test].

Now let us assume $a_n=\displaystyle\sum_{n=1}^\infty\frac {n^{n}}{e^nn!}$
Then $\frac{a_{n+1}}{a_n}=\displaystyle \dfrac {(n+1)^n}{en^n}=\displaystyle \frac {\left(1+\dfrac {1}{n}\right)^n}{e}>\displaystyle \frac {\left(1+\dfrac {1}{n}\right)^n}{\left(1+\dfrac {1}{n}\right)^{n+1}}=\displaystyle \frac {1}{\left(1+\dfrac {1}{n}\right)}=\dfrac {n}{n+1}=\dfrac {\frac{1}{n+1}}{\frac 1n}$.
Applying last lemma if $\sum a_n$ is convergent then $\sum \frac 1n$ is convergent.
Contradiction!!
