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Does the series Converge or Diverge ?

by smb26
Tags: converge, diverge, series
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smb26
#1
Mar8-12, 05:18 PM
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
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Mark44
#2
Mar8-12, 08:13 PM
Mentor
P: 21,397
Quote Quote by smb26 View Post
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
$$ \sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$
smb26
#3
Mar8-12, 11:43 PM
P: 5
ln(n) is in the deonminator

morphism
#4
Mar8-12, 11:52 PM
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?
smb26
#5
Mar9-12, 06:35 AM
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??
Mark44
#6
Mar9-12, 09:18 AM
Mentor
P: 21,397
Quote Quote by smb26 View Post
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
#7
Mar9-12, 01:17 PM
P: 5
Quote Quote by Mark44 View Post
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
#8
Mar9-12, 01:26 PM
Mentor
P: 21,397
It might or might not be helpful in the Ratio Test.
smb26
#9
Mar10-12, 08:54 AM
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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