Does the series Converge or Diverge ?

  • Thread starter smb26
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  • #1
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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

  • #2
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6,774
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
$$ \sum_{n = 2}^{\infty} \frac{ln(n!) \cdot ln(n)}{n^3}$$
 
  • #3
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ln(n) is in the deonminator
 
  • #4
morphism
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The ratio test seems to work. Care to show us how far you got along in trying to simplify?
 
  • #5
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[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??
 
  • #6
35,028
6,774
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)]
 
  • #7
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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 ???
 
  • #8
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6,774
It might or might not be helpful in the Ratio Test.
 
  • #9
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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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