Finding if a series is convergent.

[kelumhi]

Finding if a series is convergent-Answered

Homework Statement

Find for which values of K is the fallowing series convergent.

\sum((n!)²)/((kn)!)

where:
N is the variable.
K is a constant or a list of constant (eg. "(2,91]")

Homework Equations

I believe the ratio test, which states that if (f(n+1)/(f(n) as n approaches infinity is less then 1, it converges.

The Attempt at a Solution

I believe the obvious way to go about this would be the ratio test which is as fallows"

[PLAIN]http://img688.imageshack.us/img688/2140/equation1.png
[PLAIN]http://img146.imageshack.us/img146/2783/equation2.png
[PLAIN]http://img232.imageshack.us/img232/1554/equation3.png

1>((n+1)!*(n+1)!/(k(n+1))!*((Kn)!/(n!*n!) as n\rightarrow \infty

1>(n+1)(n+1)/(k(n+1)) as n\rightarrow \infty

1> (n+1)/k

which is not true, therefore this series must diverge for any possible K.

my question: am i doing anything wrong or did the teacher give a trick question?

 

[lanedance]

actually sorry, I take that back, i think there is a mistake
\frac{ (kn)! }{ (k(n+1))! } = \frac{ (kn)! }{ (kn+k)! } \neq \frac{1}{k(n+1)}

 

[lanedance]

as a guide, try k = 1,2,3 and see whether you get a converging series

another way do this covering all integer k>1 would be to use Stirlings's Approximation for n! as n gets large

 

[kelumhi]

Oh, Thanks for your help lanedance. i found it pretty intuitive that 2+ was the answer but i wouldn't have thought to be able to solve it mathematically i would need to distribute the K then simply insert a value. you've been very helpful.

 

[lanedance]

to be rigorous, you need to show its true for all k>2, not just one larger than 2 (3)

so options are
- have a look at Stirlings Approximation n! \approx n^{n}e{-n}\sqrt{2\pi n }
- try mathematical induction

i haven't tried either fully but think both should work