Help prove the limit of x^n/n = 0

[dwelch5]

Homework Statement

lim x^n/n! = 0 for all x
n->∞

Homework Equations

No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!

The Attempt at a Solution

I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?

 

[STEMucator]

Homework Statement

lim x^n/n! = 0 for all x
n->∞

Homework Equations

No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!

The Attempt at a Solution

I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?

Do you know that the factorial function grows faster than any x^n?

 

[dwelch5]

Do you know that the factorial function grows faster than any x^n?

Yeah I do know this, but I am not sure how to prove it.

 

[micromass]

Homework Statement

lim x^n/n! = 0 for all x
n->∞

Homework Equations

No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!

The Attempt at a Solution

I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?

Have you seen series?? Can you show that

\sum_{n=1}^{+\infty} \frac{x^n}{n!}

converges?

 

[Ray Vickson]

Yeah I do know this, but I am not sure how to prove it.

Assume x > 0. We then have
\log\left(\frac{x^n}{n!}\right) = n \log(x) - \log(1) - \log(2) - \cdots - \log(n).
Try to bound the sum of the logs. (Hint: compare log(j) with the integral of log(x) over [j-1,j] or [j,j+1], whichever one gives you what you need.)

RGV

 

[dwelch5]

Have you seen series?? Can you show that

\sum_{n=1}^{+\infty} \frac{x^n}{n!}

converges?

To be honest I really don't know what that means. In high school our teacher didn't worry about that kind of sigma notation because it wasn't on the AP test. So I kind of understand it after reading into it myself, but it was never anything that got set into my mind.

Assume x > 0. We then have
\log\left(\frac{x^n}{n!}\right) = n \log(x) - \log(1) - \log(2) - \cdots - \log(n).
Try to bound the sum of the logs. (Hint: compare log(j) with the integral of log(x) over [j-1,j] or [j,j+1], whichever one gives you what you need.)

RGV

I think I know what you mean. I'm going to try this out tomorrow. But thanks guys.

 

[STEMucator]

Fun fact :

e^x = \sum_{n=1}^{\infty} \frac{x^n}{n!}

 

[Ray Vickson]

Homework Statement

lim x^n/n! = 0 for all x
n->∞

Homework Equations

No equations, but I am currently in Calculus II in my first semester here at my University. Please help me prove why this limit is true!

The Attempt at a Solution

I've tried to use natural logarithms and L'Hospital's rule, but I don't know if I can derive (n!). Any ideas?

You can also use the fact that for 0 < x < N, and n > N we have
0 &lt; \frac{x^n}{n!} &lt; \frac{N^n}{n!} <br /> = \frac{N^N}{N!} \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n}.
For any r in (0,1) we have
\frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n} &lt; r^{n-N} for all n sufficiently large (why?) so the desired result follows for any x.

RGV

 

[jbunniii]

Choose any fixed x. If a_n = x^n / n!, then
\left|\frac{a_{n+1}}{a_n}\right| = \left| \frac{x^{n+1} n!}{x^n (n+1)!}\right| = \frac{|x|}{n+1}
which is arbitrarily small if n is large enough. In particular, if I choose r with 0 &lt; r &lt; 1, then for sufficiently large n, say n \geq N, we have |a_{n+1}/a_{n}| &lt; r. What can you conclude?

 

[dwelch5]

You can also use the fact that for 0 < x < N, and n > N we have
0 &lt; \frac{x^n}{n!} &lt; \frac{N^n}{n!} <br /> = \frac{N^N}{N!} \frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n}.
For any r in (0,1) we have
\frac{N}{N+1} \frac{N}{N+2} \cdots \frac{N}{n} &lt; r^{n-N} for all n sufficiently large (why?) so the desired result follows for any x.

RGV

This explanation actually helped a lot. Thanks so much to you and the others that tried helping