# [SOLVED] Use ratio test to determine if series converges or diverges

• Apr 23rd 2010, 06:50 PM
vinson24
[SOLVED] Use ratio test to determine if series converges or diverges
$\displaystyle \sum_{n=1}^{\infty}\ \frac{k!}{k^3}$
I'm stuck about half way through
this is what ive done so far
$\displaystyle lim_{k->\infty}\frac{(k+1)!}{(k+1)^3}*\frac{k^3}{k!}$
I know that getting rid of the factorials gives me k+1 in the numerator but im not sure how to simplify the $\displaystyle k^3$and $\displaystyle (k+1)^3$
• Apr 23rd 2010, 07:05 PM
chisigma
Is $\displaystyle \frac{(k+1)!}{(k+1)^{3}}\cdot \frac{k^{3}}{k!} = (k+1)\cdot (1-\frac{1}{k})^{3}$ so that...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 23rd 2010, 07:11 PM
Soroban
Hello, vinson24!

Quote:

$\displaystyle \sum_{n=1}^{\infty}\ \frac{k!}{k^3}$

I'm stuck about half way through

This is what ive done so far: .$\displaystyle \lim_{k->\infty}\frac{(k+1)!}{(k+1)^3}\cdot\frac{k^3}{k! }$

I know that getting rid of the factorials gives me k+1 in the numerator
but im not sure how to simplify the $\displaystyle k^3$ and $\displaystyle (k+1)^3$

We have: .$\displaystyle \frac{(k+1)!}{k!}\cdot\frac{k^3}{(k+1)^3} \;=\;\frac{k+1}{1}\cdot\left(\frac{k}{k+1}\right)^ 3$

In the parentheses, divide top and bottom by $\displaystyle k\!:\quad \frac{k+1}{1}\cdot\left(\frac{1}{1+\frac{1}{k}}\ri ght)^3$

Got it?

• Apr 23rd 2010, 07:29 PM
vinson24
yes diverges thanks