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

In summary, the conversation discusses a method for finding a sufficiently large "n" to satisfy a given statement. The conversation also mentions the use of the least integer and greatest integer functions, as well as the use of TeX for mathematical notation. The validity of this method is questioned and clarified.
  • #1
Derek Hart
14
1
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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  • #2
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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  • #3
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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  • #4
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
 

1. What is the definition of a limit at infinity?

A limit at infinity is the value that a function approaches as its input approaches infinity.

2. How can we prove that (x^n)/n has a limit of 0 at infinity?

This can be proven using the limit definition of infinity, which states that for any positive number M, there exists a value N such that when x is greater than N, the value of (x^n)/n is less than M.

3. What is the role of n in the limit at infinity of (x^n)/n?

The exponent n determines the rate at which the function approaches infinity. As n increases, the function approaches infinity at a slower rate, resulting in a limit of 0.

4. Are there any restrictions on the value of n for the limit at infinity of (x^n)/n to be 0?

Yes, n must be a positive integer for the limit at infinity of (x^n)/n to be 0. If n is negative, the function will approach infinity at a faster rate and the limit will not be 0.

5. How does the limit at infinity of (x^n)/n relate to other limits at infinity?

The limit at infinity of (x^n)/n is a specific case of the more general limit at infinity of (x^m)/n, where m and n are real numbers. As n approaches infinity, the value of the function approaches 0 for any value of m greater than 0.

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