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

In summary, we used the ratio test to solve the series ∞Ʃ (((n)!)^3)/(3(n))!. After rearranging the terms and taking the limit as n approaches infinity, we found that the series converges to 1/27.
  • #1
BigJon
24
0
∞
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.
 
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  • #2
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
BigJon said:
∞
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).
 
  • #4
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)
 

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  • #5
BigJon said:
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.
 
  • #6
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
BigJon said:
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?
 
  • #8
No I am getting 1/27 when i divided the highest terms
 

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

1. What is the ratio test method for evaluating infinite series?

The ratio test is a method used to determine the convergence or divergence of an infinite series. It involves taking the limit of the ratio of consecutive terms in the series. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, the test is inconclusive and another method must be used.

2. How do you apply the ratio test to the infinite series ∞ Ʃ (n!)^3/3(n)?

To apply the ratio test, we first need to rewrite the series in a form that allows us to easily take the limit. In this case, we can rewrite the series as ∞ Ʃ (n!)^3/3^n. Then, we take the limit of the absolute value of the ratio of consecutive terms, which in this case is lim|((n+1)!)^3/3^(n+1))/((n!)^3/3^n)|. If this limit is less than 1, the series converges. If it is greater than 1, the series diverges.

3. What is the significance of the factorial term in the series ∞ Ʃ (n!)^3/3(n)?

The factorial term represents the growth rate of the series. In this case, the (n!)^3 term grows much faster than the 3^n term, which means that the series is likely to converge. This is because the terms in the series are getting smaller at a faster rate, indicating convergence.

4. Can the ratio test be used to determine the exact sum of an infinite series?

No, the ratio test can only determine the convergence or divergence of an infinite series. To find the exact sum of an infinite series, other methods such as the geometric series test or telescoping series method must be used.

5. Are there any limitations to using the ratio test to evaluate infinite series?

Yes, the ratio test is only applicable to series with positive terms. It also may not work in cases where the limit of the ratio of consecutive terms is equal to 1, in which case another method must be used to determine convergence or divergence. Additionally, the ratio test does not work for alternating series or series with terms that do not decrease monotonically.

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