MHB What is the limit of x^n/n as n approaches infinity?

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The discussion centers on proving that the limit of x^n/n! approaches 0 as n approaches infinity. Various methods are suggested, including a humorous approach and a more rigorous application of the Stolz–Cesàro theorem. Some participants express skepticism about the validity of certain proofs, particularly regarding assumptions about the existence of limits. The conversation highlights the diversity of methods available for tackling this limit problem. Overall, the thread showcases a collaborative effort to explore mathematical proofs and their nuances.
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Prove that $\lim_{{n}\to{\infty}}\frac{x^n}{n!}=0$.
 
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There are so many ways to approach this but here's a funny way.

Since the series $\sum_{n = 0}^\infty x^n/n!$ converges to $e^x$ for all $x$, by the $n$th term test, $\lim_{n\to \infty} x^n/n! = 0$ for all $x$.
 
Assume that $x>0$
$$\frac{x^n}{n!} \leq \frac{k^n}{n!} \leq \frac{k^k}{k!}\times \frac{k}{n}\, \to 0 $$[/sp]
 
My solution:

By the Stolz–Cesàro theorem, we may state:

$$\lim_{n\to\infty}\frac{x^n}{n!}=\lim_{n\to\infty}\frac{x^{n+1}-x^n}{(n+1)!-n!}=\lim_{n\to\infty}\left(\frac{x^n}{n!}\cdot\frac{x-1}{n}\right)=\left(\lim_{n\to\infty}\frac{x^n}{n!}\right)\cdot0=0$$
 
Rido12 said:
Prove that $\lim_{{n}\to{\infty}}\frac{x^n}{n!}=0$.

[sp]The safest way, in the sense that You avoid possible logical vicious circles, consists of 'merely attacking' ...

$\displaystyle \frac{x^{n}}{n!} = \frac{x}{n}\ \frac{x}{n-1} ... \frac{x}{2}\ x\ (1)$

... and the limit of (1) if n tends to infinity is quite obvious...[/sp]

Kind regards

$\chi$ $\sigma$
 
Using Stirling's approximation,

$$\lim_{n\to\infty}\frac{x^n}{n!}$$

$$=\lim_{n\to\infty}\frac{x^n}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}$$

$$=\lim_{n\to\infty}\left[\frac{1}{\sqrt{2\pi n}}\cdot\left(\frac{ex}{n}\right)^n \right]=0$$
 
MarkFL said:
My solution:

By the Stolz–Cesàro theorem, we may state:

$$\lim_{n\to\infty}\frac{x^n}{n!}=\lim_{n\to\infty}\frac{x^{n+1}-x^n}{(n+1)!-n!}=\lim_{n\to\infty}\left(\frac{x^n}{n!}\cdot\frac{x-1}{n}\right)=\left(\lim_{n\to\infty}\frac{x^n}{n!}\right)\cdot0=0$$

I don't think this is a valid proof since you are assuming that the limit exists.
 
ZaidAlyafey said:
I don't think this is a valid proof since you are assuming that the limit exists.

The Stolz–Cesàro theorem asserts that if the limit:

$$\lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L$$

exists, then the limit:

$$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L$$

also exists and is equal to $L$.
 
MarkFL said:
The Stolz–Cesàro theorem asserts that if the limit:

$$\lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L$$

exists, then the limit:

$$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L$$

also exists and is equal to $L$.

For the proof of the first statement you are using the second statement which is what we want to prove.
 
  • #10
ZaidAlyafey said:
For the proof of the first statement you are using the second statement which is what we want to prove.

No, I am using the first statement to prove the second. I am using the fact that:

$$\lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L$$

exists and is equal to $L$ to assert that:

$$\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L$$
 
  • #11
ZaidAlyafey said:
For the proof of the first statement you are using the second statement which is what we want to prove.

Ahhh...my apologies for being so thick this morning...

I now (finally) see what you are saying...in my statement:

$$\lim_{n\to\infty}\frac{x^n}{n!}=\lim_{n\to\infty}\frac{x^{n+1}-x^n}{(n+1)!-n!}=\lim_{n\to\infty}\left(\frac{x^n}{n!}\cdot\frac{x-1}{n}\right)={\color{red}\left(\lim_{n\to\infty}\frac{x^n}{n!}\right)}\cdot0=0$$

I am in fact assuming the part in red exists.
 
  • #12
Another attempt:

This is in fact very similar to Euge's method...

Let:

$$S=\sum_{k=1}^{\infty}\left(a_k\right)=\sum_{k=1}^{\infty}\left(\frac{x^k}{k!}\right)$$

To determine if $S$ converges, and thus if the terms tend to zero, we may use the ratio test:

$$L=\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=\lim_{k\to\infty}\left|\frac{x^{k+1}}{(k+1)!}\cdot\frac{k!}{x^k}\right|=\lim_{k\to\infty}\left|\frac{x}{k+1}\right|=0$$

Since $L=0$, the series converges, and this implies:

$$\lim_{n\to\infty}\frac{x^n}{n!}=0$$
 
  • #13
Thanks to everyone that participated! I never knew there would be so many methods to solve this problem. :D My solution is in fact the same as Euge and Mark's, as it arose during my series and sequence homework.
 
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