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Sequence limit (factorial derivative?) 
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#1
Nov1512, 01:55 PM

P: 49

1. The problem statement, all variables and given/known data
Find the limit of the sequence given by [itex]S_{n}=\frac{n^{n}}{n!}[/itex] 2. Relevant equations [itex]lim_{n>∞}\frac{n^{n}}{n!}[/itex] 3. The attempt at a solution I know the sequence diverges, but that doesn't mean the limit is also ∞, right? 


#2
Nov1512, 02:02 PM

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RGV 


#3
Nov1512, 02:14 PM

P: 1,481

Ask yourself, which grows faster, the numerator or the denominator.



#4
Nov1512, 02:23 PM

P: 49

Sequence limit (factorial derivative?)
Thank you both. As the numerator grows faster, and it's a divergence sequence, then the limit is +∞. Now, where may I start to formally prove it?



#5
Nov1512, 02:30 PM

P: 1,481

Can you prove that [itex] n^n > n! [/itex] think about intervals.



#6
Nov1512, 03:09 PM

P: 49

Yes, I can. For instance, I just evaluated [itex]lim_{n>∞}\frac{3^{n}}{(n+3)!}[/itex] as being 0 by showing that [itex]\forall x \geq 0, (n+3)!>3^{n}[/itex]. In a similar way, I may show that [itex]n^{n}>n![/itex] in the same interval.
It just seems to me that this method isn't rigorous enough, you know? As an example, 3>1 is true, but that does't mean that [itex]lim_{x>∞}\frac{x}{3x}=0[/itex] Thank you 


#7
Nov1512, 03:15 PM

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You don't see that for arbitrary n
n times n > 1 times 2 times ... times n ? 


#8
Nov1512, 03:21 PM

P: 49

Yes, I do, but this seems as an intuitive approach, to me. Isn't it?
Just to be sure I got the concepts correctly: Let [itex]s_{n}=\frac{3^{n}}{(n+3)!}[/itex]. Then, [itex]s_{1}, s_{2}, ..., s_{n}[/itex] is a sequence, and the partial sum is [itex]S_{x}=s_{1}+s_{2}+...+s_{x}[/itex]. That being said, when I say I want to know the limit of the sequence [itex](lim_{n>∞}\frac{3^{n}}{(n+3)!})[/itex], I'm evaluating the "last" term, [itex]s_{n}[/itex], not the sum to the "last" term, [itex]S_{n}[/itex] right? Many thanks 


#9
Nov1512, 03:25 PM

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RGV 


#10
Nov1512, 03:51 PM

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Can you prove that, for n>3
[tex] n^n > \frac{n^n}{n!} > n +1 [/tex] ? 


#11
Nov1812, 08:27 PM

P: 49

Thanks 


#12
Nov1912, 10:45 AM

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I don't think my method for the second inequality (the 1st is obvious) is induction.



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