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Part 1: For what values of $k$ is the limit $<1$:

$\lim_{n\to \infty} \dfrac{(n+1)^4}{kn +k!} < 1$ where we can choose $k$ to satisfy this inequality.

How would I start? I see where $k=n$ the limit is clearly $0$ but what else can I do?

Part 2: Is this ratio test correct: $\sum \limits_1^\infty \dfrac{(n!)^4}{(kn)!}$


$\lim_{n \to \infty} \dfrac{((n+1)!)^4}{(k(n+1))!} \cdot \dfrac{(kn)!}{(n!)^4} = \dfrac{(n+1)^4(n!)^4}{(kn +k)(kn+k-1)(kn+k-2)...(kn)!} \cdot \dfrac{(kn)!}{(n!)^4}$

Which is:

$\dfrac{(n+1)^4}{(kn)^k +something + k!}$ So I would say the series converges when $k \geq 4$

  1. Is it possible to easily determine "something" or would this be your analysis?
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I can't understand question exactly... –  Detectives Nov 7 '12 at 4:37
What values of $k$ will the limit be less than one –  CodeKingPlusPlus Nov 7 '12 at 4:40
@CodeKingPlusPlus By what you have written $k$ is a fixed number. Hence, for any $k$, the limit will blow off to infinity. –  user17762 Nov 7 '12 at 4:41
Indeed, choose k for any nonconstant polynomial function of n, which diverges to infinity when n goes to infinity we have limit 0. –  Detectives Nov 7 '12 at 4:44
@mathlover Check out part2 in my edit. Part 1 came from an error carrying out the ratio test. But now I think I have it. –  CodeKingPlusPlus Nov 7 '12 at 5:11

2 Answers 2

up vote 1 down vote accepted

For part 1, all you have to do is choose $k$ so that $k!\gt(n+1)^4$. For any positive $\epsilon$, $k=n^{\epsilon}$ will do.

For part 2, you certainly have convergence for $k\gt4$, and divergence for $k\lt4$, but the case $k=4$ needs to be treated a little more delicately.

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for part 2, wouldnt the limit just be $\frac{1}{4}$ when $k = 4$? What are your thoughts about that "something"? –  CodeKingPlusPlus Nov 7 '12 at 5:22
I get $(1/4)^4$. –  Gerry Myerson Nov 7 '12 at 6:14
Yeah that is correct, and that is $< 1$ so it converges –  CodeKingPlusPlus Nov 7 '12 at 6:18

For any constant $\,k\in\Bbb N\,$,

$$\frac{(n+1)^4}{kn+k!}\xrightarrow [n\to\infty]{}\infty$$

since, for example,

$$\frac{(n+1)^4}{kn+k!}\geq\frac{(n+1)^4}{kn+n}\xrightarrow [n\to\infty]{} \infty$$

for $\,n>k!\,$

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Checkout the edit I made. The original question was motivated by an infinite series in which I incorrectly determined –  CodeKingPlusPlus Nov 7 '12 at 5:11

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