Find Positive Integer K for Convergent Series

[tnutty]

For which positive integers k is the following series convergent?
$$\sum(n!)^2/(kn)!$$attempt: use the ratio test.

let B_n = (n!)² / (kn!)

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

then B_n+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?

 

[Mark44]

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?

 

[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?

 

[Mark44]

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.

 

[tnutty]

Wait, my online homework only gives the option

k ? _____

the ? is >= , <= .

 

[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.

 

[tnutty]

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

 

[Mark44]

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.

 

[HallsofIvy]

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


attempt: use the ratio test.

let B_n = (n!)² / (kn!)

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

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

then B_n+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?

 

[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.

 

[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.

 

[tnutty]

no takers ?

 

[tnutty]

still want help if you know how to