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

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Discussion Overview

The discussion revolves around the proof that the limit of (x^n)/n approaches 0 as n approaches infinity. Participants explore different reasoning approaches, mathematical expressions, and clarifications regarding notation and definitions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a proof assuming x is positive and introduces a method involving the smallest natural number greater than x, denoted as p.
  • The proof involves rewriting the expression and establishing conditions under which (x^n/n!) is less than a given ε, leading to a requirement for n.
  • Another participant confirms the validity of the approach but points out potential ambiguity in the notation used for the denominator in the inequality.
  • A third participant introduces the concept of the least integer function (ceiling function) as a way to express the choice of p, providing an example for clarity.
  • Participants discuss the importance of clear mathematical notation and suggest using TeX for better representation of expressions.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the proof approach presented, but there are discussions regarding the clarity of notation and expressions. No consensus on the overall correctness of the proof is reached.

Contextual Notes

There are unresolved issues regarding the clarity of mathematical notation, particularly concerning the placement of terms in the denominator of the inequality. The discussion also highlights the need for precise definitions when using functions like the ceiling function.

Derek Hart
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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?
 
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That works, sure.

Derek Hart said:
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)!}##.
 
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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##.
 
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mfb said:
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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