1. ## Convergence

Hi,

I just want to make sure I've done this right.... I have to find out if this diverges or converges, so I used the ratio test and showed the limit is 1/3, meaning it converges.

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

I'm not sure how I should do this one...

$\displaystyle \sum_{n=1}^{\infty} ne^{-n}$

Thanks

2. Hello,
Originally Posted by Benno_11
Hi,

I just want to make sure I've done this right.... I have to find out if this diverges or converges, so I used the ratio test and showed the limit is 1/3, meaning it converges.

$\displaystyle \sum_{n=1}^{\infty} \frac{n!}{n^n}$
How did you find 1/3 ?

$\displaystyle \frac{a_{n+1}}{a_n}=\dots=(n+1)\cdot\frac{n^n}{(n+ 1)^{n+1}}=\left(\frac{n}{n+1}\right)^n=\left(1-\frac 1n\right)^n$

Then remember that $\displaystyle \lim_{n\to\infty}\left(1+\frac an\right)^n=e^a$

So here, the limit is $\displaystyle e^{-1}=\frac 1e<1$

I'm not sure how I should do this one...

$\displaystyle \sum_{n=1}^{\infty} ne^{-n}$

Thanks
I'm sure that this writing may enlighten you :
$\displaystyle e^{-n}=(e^{-1})^n$

3. haha, what I meant to write for the first one was
$\displaystyle \sum_{n=1}^{\infty} \frac{n^3}{3^n}$,

Thanks for the help.