Attachment 25964

can anyone explain how why (kn!) / (kn+k)! is K^K in the final part of the answer?

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- Nov 27th 2012, 05:16 PMbakinbacon(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 PMBeanRe: (kn)! / (kn+k)! = k^k?
Oops. Didn't realize it was a limit at first glance.

- Nov 27th 2012, 05:52 PMbakinbaconRe: (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 PMBeanRe: (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 PMBeanRe: (kn)! / (kn+k)! = k^k?
$\displaystyle \frac{kn}{n} + \frac{k}{n} = k$ as n goes to infinity.

- Nov 27th 2012, 06:35 PMMarkFLRe: (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:

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