1. series with factorial

$\sum_{n=1}^{\infty}\frac{n!}{n^n}$

determine whether the series converges or diverges

help?

2. Hello, buttonbear!

Without the Ratio Test, I'm not sure of the proof.

$\sum_{n=1}^{\infty}\frac{n!}{n^n}$ . Determine whether the series converges or diverges
The Ratio Test provides some interesting maniputations . . .

$R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!} \;=\;\frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n+1}}$

. . $= \;\frac{n+1}{1}\cdot\frac{n^n}{(n+1)^{n+1}} \;=\;\frac{n^n}{(n+1)^n} \;=\;\left(\frac{n}{n+1}\right)^n$

Divide top and bottom of the fraction by $n\!:\;\;\left(\frac{1}{1+\frac{1}{n}}\right)^n \;=\;\frac{1}{\left(1 + \frac{1}{n}\right)^n}$

Therefore: . $\lim_{n\to\infty}R \;=\;\lim_{n\to\infty}\frac{1}{\left(1 + \frac{1}{n}\right)^n} \;=\;\frac{1}{\lim\left(1 + \frac{1}{n}\right)^n} \;=\;\frac{1}{e} \;<\;1$

. . and the series converges.

3. For all $n \geq 2$, we have: $\frac{n!}{n^n} = \frac{1 \cdot 2 \cdot 3 \cdots (n-1)n}{n \cdot n \cdot n \cdots n \cdot n} \ \ {\color{red} \leq}\ \ \frac{1}{n} \cdot \frac{2}{n} \cdot 1 \cdot 1 \cdots \cdot 1 = \frac{2}{n^2}$

So by comparison ...

4. i'm sorry, i just didn't fully understand your post..

5. I've showed that: $\frac{n!}{n^n} < \frac{2}{n^2}$

Since $\sum \frac{2}{n^2}$ is convergent, by the comparison test ... what can you conclude?

6. i can conclude that... because the larger sequence converges..the other one does too?

7. i'm just kind of confused as how to how you knew to pick $\frac{2}{n^2}$ for your larger sequence? or can it be any larger sequence?

8. sorry for bumping up an old thread but how did you know to compare it to 2/n^2 as opposed to 1/n^2??

9. oh nvm i think i see it, because you cant use 1/n^2, you can only use 2/n^2, cause if you only use the first term it would be 1/n, which is divergent according to the p-series, and so you use n>=2..ahh i see it..