Determine whether the following series converges

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The discussion centers on determining the convergence of the series involving ln(n) and n!. The Ratio Test is applied, leading to a transformation from step 1 to step 2, where the factorials are simplified. It is clarified that n!/(n + 1)! equals 1/(n + 1), which aids in understanding the convergence behavior. The inequality e^n > n is used to establish that ln(n) grows slower than n!, supporting the conclusion that the series converges. Ultimately, the series ln(n)/n! converges due to its comparison with the convergent series 1/(n-1)!.
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Could someone please explain how they got from step 1 to step 2, i don't know what hap'd to the factorial part :S

Question: Determine whether the following series converges.

n= infinity
Sum ln (n)
n=1 n!

Using the Ratio Test: lim absolute ((An+1)/(An))
n->infinity

Step 1. => lim (ln(n+1)/(n+1)!) x (n!/(ln (n))

Step 2. => lim ln(n+1)/((n+1)ln(n))


Link to solution: http://www.maths.uq.edu.au/courses/MATH1051/Semester2/Tutorials/prob9sol.pdf

It's question 1. vi)
 
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n!/(n + 1)! = 1/(n + 1).

If you write out the factorials for n = 1, 2, 3 you should see why this is the case.
 
Using this equality e^x >= x + 1 > x for all x, we have e^n > n for all n. Thus, ln(n) < n for all n. This implies 0 < ln(n)/n! < 1/(n-1)! for all n>1
The series (Sum 1/(n-1)!) is convergent, so is the series (Sum ln(n)/n!)
 
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