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Help prove the limit of x^n/n! = 0

  • Thread starter dwelch5
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  • #1
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Homework Statement


lim x^n/n! = 0 for all x
n->∞


Homework Equations


No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!


The Attempt at a Solution


I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?
 

Answers and Replies

  • #2
Zondrina
Homework Helper
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Homework Statement


lim x^n/n! = 0 for all x
n->∞


Homework Equations


No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!


The Attempt at a Solution


I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?
Do you know that the factorial function grows faster than any x^n?
 
  • #3
4
0
Do you know that the factorial function grows faster than any x^n?
Yeah I do know this, but I am not sure how to prove it.
 
  • #4
22,097
3,277

Homework Statement


lim x^n/n! = 0 for all x
n->∞


Homework Equations


No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!


The Attempt at a Solution


I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?
Have you seen series?? Can you show that

[tex]\sum_{n=1}^{+\infty} \frac{x^n}{n!}[/tex]

converges?
 
  • #5
Ray Vickson
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Yeah I do know this, but I am not sure how to prove it.
Assume x > 0. We then have
[tex] \log\left(\frac{x^n}{n!}\right) = n \log(x) - \log(1) - \log(2) - \cdots - \log(n).[/tex]
Try to bound the sum of the logs. (Hint: compare log(j) with the integral of log(x) over [j-1,j] or [j,j+1], whichever one gives you what you need.)

RGV
 
  • #6
4
0
Have you seen series?? Can you show that

[tex]\sum_{n=1}^{+\infty} \frac{x^n}{n!}[/tex]

converges?
To be honest I really don't know what that means. In high school our teacher didn't worry about that kind of sigma notation because it wasn't on the AP test. So I kind of understand it after reading into it myself, but it was never anything that got set into my mind.

Assume x > 0. We then have
[tex] \log\left(\frac{x^n}{n!}\right) = n \log(x) - \log(1) - \log(2) - \cdots - \log(n).[/tex]
Try to bound the sum of the logs. (Hint: compare log(j) with the integral of log(x) over [j-1,j] or [j,j+1], whichever one gives you what you need.)

RGV
I think I know what you mean. I'm going to try this out tomorrow. But thanks guys.
 
  • #7
Zondrina
Homework Helper
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Fun fact :

[tex]e^x = \sum_{n=1}^{\infty} \frac{x^n}{n!}[/tex]
 
  • #8
Ray Vickson
Science Advisor
Homework Helper
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1,722

Homework Statement


lim x^n/n! = 0 for all x
n->∞


Homework Equations


No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!


The Attempt at a Solution


I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?
You can also use the fact that for 0 < x < N, and n > N we have
[tex]0 < \frac{x^n}{n!} < \frac{N^n}{n!}
= \frac{N^N}{N!} \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n}. [/tex]
For any r in (0,1) we have
[tex] \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n} < r^{n-N} [/tex] for all n sufficiently large (why?) so the desired result follows for any x.

RGV
 
  • #9
jbunniii
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Choose any fixed [itex]x[/itex]. If [itex]a_n = x^n / n![/itex], then
[tex]\left|\frac{a_{n+1}}{a_n}\right| = \left| \frac{x^{n+1} n!}{x^n (n+1)!}\right| = \frac{|x|}{n+1}[/tex]
which is arbitrarily small if [itex]n[/itex] is large enough. In particular, if I choose [itex]r[/itex] with [itex]0 < r < 1[/itex], then for sufficiently large [itex]n[/itex], say [itex]n \geq N[/itex], we have [itex]|a_{n+1}/a_{n}| < r[/itex]. What can you conclude?
 
  • #10
4
0
You can also use the fact that for 0 < x < N, and n > N we have
[tex]0 < \frac{x^n}{n!} < \frac{N^n}{n!}
= \frac{N^N}{N!} \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n}. [/tex]
For any r in (0,1) we have
[tex] \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n} < r^{n-N} [/tex] for all n sufficiently large (why?) so the desired result follows for any x.

RGV
This explanation actually helped a lot. Thanks so much to you and the others that tried helping
 

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