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

AI Thread Summary
The discussion focuses on a proof that (x^n)/n approaches 0 as n approaches infinity, with the author presenting a unique reasoning method. They define p as the smallest natural number greater than x and rewrite the expression to analyze its behavior as n increases. The proof suggests that for sufficiently large n, the expression can be made less than any ε>0 by appropriately choosing n. Participants clarify the notation used, particularly regarding the denominator, to avoid ambiguity. Overall, the reasoning is deemed valid, with suggestions for clearer mathematical representation.
Derek Hart
Messages
14
Reaction score
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?
 
Mathematics news on Phys.org
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)!}##.
 
  • Like
Likes Derek Hart and scottdave
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##.
 
  • Like
Likes Derek Hart
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top