Does the series Converge or Diverge ?

[smb26]

1. Homework Statement

∞
Ʃ ln(n!)/(n^3)*ln(n)
n=2

2. Homework 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

 

[Mark44]

1. Homework Statement

∞
Ʃ ln(n!)/(n^3)*ln(n)
n=2

2. Homework 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
$$ \sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$

 

[smb26]

ln(n) is in the deonminator

 

[morphism]

The ratio test seems to work. Care to show us how far you got along in trying to simplify?

 

[smb26]

[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??

 

[Mark44]

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)]

 

[smb26]

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 ?

 

[Mark44]

It might or might not be helpful in the Ratio Test.

 

[smb26]

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