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- Nov 27th 2012, 05:16 PM #1

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- Nov 27th 2012, 05:37 PM #2

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- Nov 27th 2012, 05:52 PM #3

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- Nov 27th 2012, 06:23 PM #4

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## 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 #5

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- Nov 27th 2012, 06:35 PM #6
## 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:

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