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

**bakinbacon** · November 27th 2012, 05:16 PM

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

**Bean** · November 27th 2012, 05:37 PM

Oops. Didn't realize it was a limit at first glance.

**bakinbacon** · November 27th 2012, 05:52 PM

i just don't know how to write out (kn+k)!.. she hardly taught us how to do factorials

**Bean** · November 27th 2012, 06:23 PM

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.

**Bean** · November 27th 2012, 06:28 PM

$\frac{kn}{n} + \frac{k}{n} = k$ as n goes to infinity.

**MarkFL** · November 27th 2012, 06:35 PM

Within the absolute value bars, disregarding the $x$ what you have are $k$ th degree polynomials in the numerator and denominator in $n$ , so we know the limit as $n\to\infty$ is the ratio of the leading coefficients, which is:

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

And so the limit of the entire expression is:

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

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

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Re: (kn)! / (kn+k)! = k^k?

Re: (kn)! / (kn+k)! = k^k?

Re: (kn)! / (kn+k)! = k^k?

Re: (kn)! / (kn+k)! = k^k?