Sequence limit (factorial derivative?)


by carlosbgois
Tags: derivative, factorial, limit, sequence
carlosbgois
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#1
Nov15-12, 01:55 PM
P: 47
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?
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Ray Vickson
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#2
Nov15-12, 02:02 PM
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Quote Quote by carlosbgois View Post
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?
Either the function f(n) = n^n / n! converges, or else f(n) → +∞ or f(n) → -∞ or else f(n) "oscillates" as n → ∞ in such a way that f(n) does not approach a definite value---not even ∞. You need to decide which applies here.

RGV
Zondrina
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#3
Nov15-12, 02:14 PM
P: 1,338
Ask yourself, which grows faster, the numerator or the denominator.

carlosbgois
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#4
Nov15-12, 02:23 PM
P: 47

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?
Zondrina
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#5
Nov15-12, 02:30 PM
P: 1,338
Can you prove that [itex] n^n > n! [/itex] think about intervals.
carlosbgois
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#6
Nov15-12, 03:09 PM
P: 47
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
dextercioby
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#7
Nov15-12, 03:15 PM
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You don't see that for arbitrary n

n times n > 1 times 2 times ... times n ?
carlosbgois
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#8
Nov15-12, 03:21 PM
P: 47
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
Ray Vickson
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#9
Nov15-12, 03:25 PM
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Quote Quote by Zondrina View Post
Can you prove that [itex] n^n > n! [/itex] think about intervals.
This is not quite enough: you need [itex] n^n / n! [/itex] to be unbounded, not just > 1.

RGV
dextercioby
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#10
Nov15-12, 03:51 PM
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Can you prove that, for n>3

[tex] n^n > \frac{n^n}{n!} > n +1 [/tex] ?
carlosbgois
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#11
Nov18-12, 08:27 PM
P: 47
Quote Quote by dextercioby View Post
Can you prove that, for n>3

[tex] n^n > \frac{n^n}{n!} > n +1 [/tex] ?
May it be done by induction? It clearly holds for n=3, then I assume it also holds for n=j, and show it's also valid for n=j+1. (Sorry, no paper and pen around right now, I'll try it as soon as I can)

Thanks
dextercioby
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#12
Nov19-12, 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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