# (kn)! / (kn+k)! = k^k?

• Nov 27th 2012, 05:16 PM
bakinbacon
(kn)! / (kn+k)! = k^k?
Attachment 25964

can anyone explain how why (kn!) / (kn+k)! is K^K in the final part of the answer?
• Nov 27th 2012, 05:37 PM
Bean
Re: (kn)! / (kn+k)! = k^k?
Oops. Didn't realize it was a limit at first glance.
• Nov 27th 2012, 05:52 PM
bakinbacon
Re: (kn)! / (kn+k)! = k^k?
i just don't know how to write out (kn+k)!.. she hardly taught us how to do factorials
• Nov 27th 2012, 06:23 PM
Bean
Re: (kn)! / (kn+k)! = k^k?
In the final part of the ratio test, divide the numerator and the denominator by n and take n to infinity which takes all of the constants to zero. The bottom simplifies to k*k for k times. The numerator simplifies to 1^(k) or 1.
• Nov 27th 2012, 06:28 PM
Bean
Re: (kn)! / (kn+k)! = k^k?
$\displaystyle \frac{kn}{n} + \frac{k}{n} = k$ as n goes to infinity.
• Nov 27th 2012, 06:35 PM
MarkFL
Re: (kn)! / (kn+k)! = k^k?
Within the absolute value bars, disregarding the $\displaystyle x$ what you have are $\displaystyle k$th degree polynomials in the numerator and denominator in $\displaystyle n$, so we know the limit as $\displaystyle n\to\infty$ is the ratio of the leading coefficients, which is:

$\displaystyle \frac{1}{k^k}$

And so the limit of the entire expression is:

$\displaystyle \left|\frac{x}{k^k} \right|$