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Does the series Converge or Diverge ? 
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#1
Mar812, 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 


#2
Mar812, 08:13 PM

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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
Mar812, 11:43 PM

P: 5

ln(n) is in the deonminator



#4
Mar812, 11:52 PM

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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?



#5
Mar912, 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?? 


#6
Mar912, 09:18 AM

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ln(n!)/[(n^3)*ln(n)] 


#7
Mar912, 01:17 PM

P: 5




#8
Mar912, 01:26 PM

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P: 21,280

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



#9
Mar1012, 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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