Proof that (x^n)/n! has a limit of 0 at infinity

  • #1
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1

Main Question or Discussion Point

I understand that the standard proof is a bit different from my own, but I want to know if my reasoning is valid. PROOF:
Firstly, I assume that x is positive.

I then consider p = inf{n∈ℕ : n>x} . In other words, I choose "p" to be the smallest natural number greater than x. If we choose n>p, then we can rewrite the original statement as ((x/1)(x/2)...(x/(p-1)))*((x/p)...(x/n)). If we let α=((x/1)(x/2)...(x/(p-1))), then xn/n! = α((x/p)...(x/n)) where each term x/k is less than one. Thus α((x/p)...(x/n)) < α(x/n). So if we want to require that (xn/n!) < ε for some ε>0, then we can choose n satisfying α(x/n) < ε ⇒ n > (αx/ε). Since α = (xp-1/(p-1)!), we must have n > (xp/ε(p-1)!). Thus, any n satisfying n > max[p, (xp/ε(p-1)!)] will ensure that xn/n! < ε.

Is this subsequent method for finding sufficiently large "n" viable in practice, and are my findings valid?
 

Answers and Replies

  • #2
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That works, sure.

xp/ε(p-1)!
This could be misinterpreted. What exactly is part of the denominator?
Better write it as xp/(ε(p-1)!) to avoid ambiguity.
Or, even better, use TeX: ##\displaystyle n > \frac{x^p}{\epsilon (p-1)!}##.
 
  • #3
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Derek Hart said:
I then consider p = inf{n∈ℕ : n>x} . In other words, I choose "p" to be the smallest natural number greater than x.
There's a function that does this, the least integer function, also known as the ceiling function. See http://www.mathwords.com/c/ceiling_function.htm. The notation is ##\lceil x \rceil##. Unrendered, this looks like \lceil x \rceil.

For example, ##\lceil 4.52 \rceil = 5##.

There's a related function, the greatest integer function (or floor function), denoted as ##\lfloor x \rfloor##.
 
  • #4
14
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That works, sure.

This could be misinterpreted. What exactly is part of the denominator?
Better write it as xp/(ε(p-1)!) to avoid ambiguity.
Or, even better, use TeX: ##\displaystyle n > \frac{x^p}{\epsilon (p-1)!}##.
Yeah sorry I meant for both epsilon and the factorial to be in the denominator. Didn't even notice my mistake at first
 

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