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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: $\sum \limits_1^\infty \dfrac{(n!)^4}{(kn)!}$

Compute:

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

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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    @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.2012-11-07

2 Answers 2

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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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    Yeah that is correct, and that is < 1 so it converges2012-11-07
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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 determined2012-11-07