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

[Dethrone]

Prove that $\lim_{{n}\to{\infty}}\frac{x^n}{n!}=0$.

 

[Euge]

There are so many ways to approach this but here's a funny way.

Spoiler

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 $x$.

 

[alyafey22]

Spoiler

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]

 

[MarkFL]

My solution:

Spoiler

By the Stolz–Cesàro theorem, we may state:

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

 

[chisigma]

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

$\chi$ $\sigma$

 

[Greg]

Spoiler

Using Stirling's approximation,

\(\displaystyle \lim_{n\to\infty}\frac{x^n}{n!}\)

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

\(\displaystyle =\lim_{n\to\infty}\left[\frac{1}{\sqrt{2\pi n}}\cdot\left(\frac{ex}{n}\right)^n \right]=0\)

 

[alyafey22]

My solution:

Spoiler

By the Stolz–Cesàro theorem, we may state:

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

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

 

[MarkFL]

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:

\(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L\)

exists, then the limit:

\(\displaystyle \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L\)

also exists and is equal to $L$.

 

[alyafey22]

The Stolz–Cesàro theorem asserts that if the limit:

\(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L\)

exists, then the limit:

\(\displaystyle \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L\)

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.

 

[MarkFL]

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:

\(\displaystyle \lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L\)

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

\(\displaystyle \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L\)

 

[MarkFL]

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...

Spoiler

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

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

I am in fact assuming the part in red exists.

 

[MarkFL]

Another attempt:

Spoiler

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

Let:

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

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

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

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

\(\displaystyle \lim_{n\to\infty}\frac{x^n}{n!}=0\)

 

[Dethrone]

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.