# Thread: convergent series

1. ## convergent series

For which positive integers k is the following series convergent?

[(n!)^2]/[(kn)!] as n goes from 1 to infinity. Please show the steps using the ratio test.

2. Originally Posted by Taurus3
For which positive integers k is the following series convergent?
[(n!)^2]/[(kn)!] as n goes from 1 to infinity. Please show the steps using the ratio test.
$\dfrac{{\left[ {\left( {n + 1} \right)!} \right]^2 }}
{{\left( {kn + k} \right)!}}\dfrac{{\left( {kn} \right)!}}
{{\left( {n!} \right)^2 }} = \dfrac{{\left( {n + 1} \right)^2 \left( {n!} \right)^2 }}
{{\left( {n!} \right)^2 }}\dfrac{{\left( {kn} \right)!}}
{{\left( {kn + k} \right)!}}$

3. I don't think that's right. Plus what are the possible values for k?

4. Why don't you think it is correct?
Do you understand the ratio test.
You finish the algebra.

5. doesn't (kn)! get cancelled since (kn + k)! = (kn)! (kn + k) (kn + (k-1)) (kn + (k-2)) + ....so on? So doesn't that just leave us with (n+2)^2 on the numerator?

6. If $k=1$ we get $\dfrac{1}{n+1}$

If $k=2$ we get $\dfrac{(2n)!}{(2n+2)!}=\dfrac{1}{(2n+2)(2n+1)}$

If $k=3$ we get $\dfrac{(3n)!}{(3n+3)!}=\dfrac{1}{(3n+3)(3n+2)(3n+1 }$

Get it?

7. but what happened to the (n + 1)^2 in the numerator? It's still a bit confusing. So what are all the possible values for k that makes this convergent?

8. Look, I am not going to do this problem for you.
So the answer to your question is: I left that factor out.
You must do somethings for yourself.
I am now done with this thread.

9. oh, thanks. Sorry.