Does the series Converge or Diverge ?

1. The problem statement, all variables and given/known data

Ʃ ln(n!)/(n^3)*ln(n)
n=2

2. Relevant equations

Ratio test, which states:
Lim as n tends to infinity of |(An+1)/An|=ρ would lead to convergent series if the limit is less than 1
it diverges if ρ is greater than 1
test is inconclusive if ρ=1

3. The attempt at a solution
I thought of applying Ratio test but didn't know how to cancel out terms
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Mentor
 Quote by smb26 1. The problem statement, all variables and given/known data ∞ Ʃ ln(n!)/(n^3)*ln(n) n=2 2. Relevant equations Ratio test, which states: Lim as n tends to infinity of |(An+1)/An|=ρ would lead to convergent series if the limit is less than 1 it diverges if ρ is greater than 1 test is inconclusive if ρ=1 3. The attempt at a solution I thought of applying Ratio test but didn't know how to cancel out terms
n! = 2 * 3 * 4 * ... * (n - 1) * n

ln(a*b) = ln(a) + ln(b)

Is ln(n) in the numerator or denominator? The way you write the series, most would read as
$$\sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$
 ln(n) is in the deonminator

Recognitions:
Homework Help

Does the series Converge or Diverge ?

The ratio test seems to work. Care to show us how far you got along in trying to simplify?
 [ln((n+1)!)]/[ln(n+1)*(n+1)^3] * [ln(n)*n^3]/[ln(n!)] now at infinity n^3/(n+1)^3 will give 1 & ln(n)/ln(n+1) will give 1 I guess but what about ln((n+1)!)/ln(n!) what is this going to give & how to prove... Can this be done by Direct Comparison Test or Limit Comparison Test??

Mentor
 Quote by smb26 ln(n) is in the deonminator
Then you need to write the expression with more grouping symbols, like so:
ln(n!)/[(n^3)*ln(n)]

 Quote by Mark44 n! = 2 * 3 * 4 * ... * (n - 1) * n ln(a*b) = ln(a) + ln(b) Is ln(n) in the numerator or denominator? The way you write the series, most would read as $$\sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$
The separation of ln(n!) will help me how ???
 Mentor It might or might not be helpful in the Ratio Test.
 I guess I found the solution fact is ##X! < X^x## so we say that by Direct Comparison Test $$\frac{ln(n!)}{n^3 *ln(n)} < \frac {ln(n^n)}{n^3 *ln(n)} = \frac {n*ln(n)}{n^3 *ln(n)} = \frac {1}{n^2 }$$ since $$\frac {1}{n^2 }$$ is a convergent series of p=2, the $$Ʃ\frac{ln(n!)}{n^3 *ln(n)}$$ converges by Direct comparison test

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