# Homework Help: Ratio Test Problem

1. Nov 7, 2012

### BigJon

Ʃ (((n)!)^3)/(3(n))! Use the ratio test to solve
n=1

So first i put it into form of (n!)^3/3n!, then applied ratio test.

from ratio got ((n+1)!)^3/(3n+1)! times (3n)!/(n!)^3

from here im on shaky ground

i go reduce the terms to (n!)^3(n+1)^3/(3n!)(3n+1) times (3n!)/(n!)^3

i then end with lim n->∞ (n+1)^3/(3n+1)

My main two questions are did i reduce terms correctly and cancel, and how to take the limit of the remaining term.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 7, 2012

### BigJon

Forgot to mention on my calculator it says the limit goes to infinity but i need to know how to actually show steps, etc

3. Nov 7, 2012

### Dick

You are being a bit too sloppy. There is a difference between 3n+1 and 3(n+1).

4. Nov 7, 2012

### BigJon

Sorry i attached a pic of the actual prob if that would help

Also i dont see where i have that problem i took (3n+1)! to (3n!)(3n+1)

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5. Nov 7, 2012

### Dick

You do have that problem. (3n+1)!/(3n)! is (3n+1). (3(n+1))!/(3n)! is not the same thing.

6. Nov 7, 2012

### BigJon

Okay i worked it again and tried to follow what you said hopefully

I attached a pic of my work, srry for the sloppiness :P

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7. Nov 7, 2012

### Dick

Yeah, I think you've got it. The limit isn't infinity, right?

8. Nov 7, 2012

### BigJon

No im getting 1/27 when i divided the highest terms