# Does the series Converge or Diverge ?

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

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## Answers and Replies

Mark44
Mentor
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
$$\sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$

ln(n) is in the deonminator

morphism
Science Advisor
Homework Helper
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??

Mark44
Mentor
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)]

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
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