# Does the series Converge or Diverge ?

by smb26
Tags: converge, diverge, series
 P: 5 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
Mentor
P: 20,409
 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
$$\sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$
 P: 5 ln(n) is in the deonminator
Sci Advisor
HW Helper
P: 2,020

## 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?
 P: 5 [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
P: 20,409
 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)]
P: 5
 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 P: 20,409 It might or might not be helpful in the Ratio Test.
 P: 5 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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