- #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?
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?