Solve Ratio Test Problem: ∞ Ʃ (n!)^3/3(n)!

[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 I am 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.

 

[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

 

[Dick]

∞
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 I am 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.

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

 

[BigJon]

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

Also i don't see where i have that problem i took (3n+1)! to (3n!)(3n+1)

 

[Dick]

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

Also i don't see where i have that problem i took (3n+1)! to (3n!)(3n+1)

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

 

[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

 

[Dick]

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

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

 

[BigJon]

No I am getting 1/27 when i divided the highest terms