Question

determining series convergence

Answer 1

\[\sum_{n=1}^{\infty} \frac{ n! }{ n^{n} }\]

Answer 2

my guess is the ratio test

Answer 3

alright so \[\lim_{n \rightarrow \infty}\frac{ n! }{ n^{n} }\] and use l'hopital

Answer 4

you know how to use it?
no not l'hopital, the ratio test

Answer 5

check that \[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}<1\]

Answer 6

for your case that means
\[\lim_{n\to \infty}\frac{(n+1)!}{(n+1)^{n+1}}\times \frac{n^n}{n!}<1\]

Answer 7

I'm not sure how to do this

Answer 8

simply evaluate that limit

\[\begin{align}\lim_{n\to\infty}\frac{{a}_{n+1}}{{a}_{n}} &=\lim_{n\to\infty}\dfrac{(n+1)!}{(n+1)^{n+1}} \cdot \dfrac{n^n}{n!} \\~\\&=?\end{align}\]

Answer 9

try to simiplify that by canceling whatever you can

Answer 10

I dont see the algebra though. What cancels?