# Check my work for.

1. Mar 27, 2009

### tnutty

For which positive integers k is the following series convergent?
$$\sum(n!)^2/(kn)!$$

attempt: use the ratio test.

let Bn = (n!)2 / (kn!)

let Bn+1 = ((n+1)!)^2 / (k(n+1))!

then Bn+1 / Bn =

((n+1)!)^2 / (k(n+1))! * (kn)!/(n!)^2

from manipulation I got :

(n+1)2 / k(n+1)

= (n+1) / k

if this is right (can you check please) the how would it follow to determine K such that
the following series is convergent?

Last edited: Mar 28, 2009
2. Mar 28, 2009

### Staff: Mentor

Your answer is right, but needs parentheses: (n + 1)/k. As you wrote it, it would be interpreted as n + (1/k), and I don't think that's what you meant.

For the ratio test to show convergence, it must be true that
$$\lim_{n \rightarrow \infty} \frac{n + 1}{k} < 1$$

Is that going to happen here?

3. Mar 28, 2009

### tnutty

I guess not. Because for any number k, if you take the limit, then the limit
of the numerator is going to be infinity, and the denominator is just going to make it still
infinity. How would you prove this deductively?

4. Mar 28, 2009

### Staff: Mentor

You've pretty much said what needs to be said. The idea is that k is some fixed number, so eventually, no matter how big k is, n will surpass it, and your ratio will be larger than 1.

5. Mar 28, 2009

### tnutty

Wait, my online hw only gives the option

k ? _____

the ? is >= , <= .

6. Mar 28, 2009

### tnutty

I don't think my work was correct. let me try again.

Root test:

$$\frac{((n+1)!)^2}{(k(n+1)!)^2}$$ x $$\frac{(kn)!}{(n!)^2}$$

=

$$\frac{(n+1)^2 * (n!)^2}{k(n+1) * (kn)!}$$ x $$\frac{(kn)!}{(n!)^2}$$

=

the kn! cancels out and the and the (n!)^2 cancels out

and we have

$$\frac{(n+1)^2}{k(n+1)}$$ < 1

(n+1)2 < k(n+1)

(n+1) < k

same as before.

maybe we should take a different approach.

just from induction, how about k being (n+1), that would make the bottom denominator
greater than the top.

7. Mar 28, 2009

### tnutty

Just guessed and the answer is 2. Can you make out why, and not just from plugging in numbers

Last edited: Mar 28, 2009
8. Mar 29, 2009

### Staff: Mentor

I think you might not have given us all the information.

Is this the series?
$$\sum_{k = 0}^\infty \frac{(n!)^2}{(kn)!}$$

You didn't include the index in your post, and I was assuming the index was n. If it's k, that's a different story.

9. Mar 29, 2009

### HallsofIvy

Staff Emeritus

Caution: (kn!) is not the same as (kn)!

10. Mar 29, 2009

### tnutty

Nope the index is n=1 to infinity. And for HallsofIvy, I meant to write (kn)! and not kn!.

Well if I look at the series I can see why the answer is 2 because the denominator will also be n^2.

11. Mar 31, 2009

### tnutty

I think this move is invalid :

$$\frac{((n+1)!)^2}{(k(n+1)!)^2}$$

=>

$$\frac{(n+1)^2 * (n!)^2}{k(n+1) * (kn)!}$$

changing the (k(n+1))! to
k(n+1) * (kn)!

This has to invalid, otherwise it does not makes any sense.

12. Apr 6, 2009

### tnutty

no takers ?

13. Apr 8, 2009

### tnutty

still want help if you know how to