Does the series $\sum n!/n^n$ converge or diverge? so I used the root test, but i'm not quite sure if i'm allowed to. I think im performing the operations correctly,a and i keep ending up with $(1)^{\infty}$. So really my question is am i performing the operations wrong or do i have to use a different test?
 A: Here is a completely elementary proof
with an explicit upper bound of 4:
$n! = \prod_{i=1}^n i$,
so, reversing the order of the terms,
$n! = \prod_{i=1}^n (n+1-i)$.
Multiplying these,
$n!^2 = \prod_{i=1}^n i(n+1-i)$.
But 
$\begin{align}
i(n+1-i) = i(n+1) - i^2
&= \frac{(n+1)^2}{4} - \frac{(n+1)^2}{4} +i(n+1) - i^2\\
&= \frac{(n+1)^2}{4} - (i-\frac{n+1}{2})^2\\
&\le \frac{(n+1)^2}{4}
\end{align}
$
so
$\begin{align}
n!^2 &\le \prod_{i=1}^n {(n+1)^2}{4}\\
&= \big(\frac{(n+1)^2}{4}\big)^n\\
&= \big(\frac{n+1}{2}\big)^{2n}\\
\end{align}
$
or
$n! \le \big(\frac{n+1}{2}\big)^{n}$,
so
$$\frac{n!}{n^n} \le \big(\frac{n+1}{2n}\big)^n
= \big(\frac1{2}+\frac1{2n}\big)^n$$
Since $\frac1{2}+\frac1{2n} \le \frac{3}{4}$ for $n \ge 2$,
$\begin{align}
\sum_{n=1}^{\infty}  \frac{n!}{n^n} 
&\le 1 + \sum_{n=2}^{\infty} \big(\frac{3}{4}\big)^n\\
&=1 + \frac{(3/4)^2}{1-3/4}\\
 &= 1 + 9/4\\
 &= 13/4\\
 &< 4\\
\end{align}
$.
A: Hint:
$$\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{1}{\left(1+\frac{1}{n}\right)^n}\xrightarrow[n\to\infty]{}\ldots$$
And yes: it converges.
A: All you need is that $n! \approx c \sqrt{n}(n/e)^n$
for some real $c$.
Then $n!/n^n \approx c \sqrt{n}/e^n$
so the sum converges.
Actually, this shows that
$\sum_{n=0}^{\infty} x^n n!/n^n$
converges for $|x| < e$.
A: Use d'Alembert's ratio test, it's not hard to find the convergence of the series. http://en.wikipedia.org/wiki/Ratio_test
A: $$\frac{n!}{n^n}=\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdot ... \cdot \frac{n}{n}\leq 
\frac{1}{n}\cdot\frac{2}{n}\cdot1\cdot ... \cdot 1= \frac{2}{n^2}$$
