Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
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
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.

share|cite|improve this answer
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 $\,n>k!\,$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.