[SOLVED] Pointwise and Uniform Convergence of x^n e^{-nx}

• Jan 25th 2010, 03:53 PM
lvleph
[SOLVED] Pointwise and Uniform Convergence of x^n e^{-nx}
There is a thread similar to this already in existence, but it is quite old and it didn't help me very much.
I need to determine the convergence of
$\lim_{n \to \infty} x^n e^{-nx}$
for $x \in [0,1]$ and $x \in [0, +\infty)$
Trying to use l'Hopital's Rule I get
$\lim_{n \to \infty} x^n e^{-nx} = \lim_{n \to \infty} \frac{x^n}{e^{nx}}$
$\lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{\log(x) x^{n-1}}{e^{nx}}$
Using l'Hopital n time results in
$\lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{(\log x)^n}{e^{nx}}$
However, I am not sure this even makes sense, because $f_n$ is not continuous with respect to n. So, what am I suppose to do?
• Jan 25th 2010, 07:12 PM
Prove It
Quote:

Originally Posted by lvleph
There is a thread similar to this already in existence, but it is quite old and it didn't help me very much.
I need to determine the convergence of
$\lim_{n \to \infty} x^n e^{-nx}$
for $x \in [0,1]$ and $x \in [0, +\infty)$
Trying to use l'Hopital's Rule I get
$\lim_{n \to \infty} x^n e^{-nx} = \lim_{n \to \infty} \frac{x^n}{e^{nx}}$
$\lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{\log(x) x^{n-1}}{e^{nx}}$
Using l'Hopital n time results in
$\lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{(\log x)^n}{e^{nx}}$
However, I am not sure this even makes sense, because $f_n$ is not continuous with respect to n. So, what am I suppose to do?

Is this a series or a sequence?
• Jan 25th 2010, 07:15 PM
lvleph
It is a sequence.

What I have to do is determine Piecewise and Uniform convergence on those intervals. I can get the Uniform once I get the Piecewise convergence. But, I can't get past that.
• Jan 25th 2010, 08:06 PM
Drexel28
Quote:

Originally Posted by lvleph
There is a thread similar to this already in existence, but it is quite old and it didn't help me very much.
I need to determine the convergence of
$\lim_{n \to \infty} x^n e^{-nx}$
for $x \in [0,1]$ and $x \in [0, +\infty)$
Trying to use l'Hopital's Rule I get
$\lim_{n \to \infty} x^n e^{-nx} = \lim_{n \to \infty} \frac{x^n}{e^{nx}}$
$\lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{\log(x) x^{n-1}}{e^{nx}}$
Using l'Hopital n time results in
$\lim_{n \to \infty} \frac{x^n}{e^{nx}} \stackrel{H}{=} \frac{(\log x)^n}{e^{nx}}$
However, I am not sure this even makes sense, because $f_n$ is not continuous with respect to n. So, what am I suppose to do?

It is clearly pointwise convergent. Now, do you think it's uniformly convergent? Will your choice of $N$ in the definition of convergence depend on $x$?
• Jan 26th 2010, 05:09 AM
lvleph
Like I said I don't need help with the Uniform Convergence. If it is clearly Pointwise then it should be easy to prove, however I am unable to figure out how.
• Jan 26th 2010, 06:54 AM
lvleph
So, I have a bit of an idea, but can't seem to figure it out.
$\lim_{n \to \infty} \left(\frac{x}{e^x}\right)^n$
This will converge to zero if $\left(\frac{x}{e^x}\right)^n < 1$ and will converge to one if $\left(\frac{x}{e^x}\right)^n = 1$, but diverges if $\left(\frac{x}{e^x}\right)^n > 1$
• Jan 26th 2010, 07:03 PM
Jose27
Consider the function $f(x)=\ln (x) -x$ then $f'(x)= \frac{1}{x} -1$ and thus $f$ is increasing on $(0,1)$ and decreasing on $(1, \infty)$ and so $f(x)\leq f(1)=-1$ for all $x\in (0, \infty)$. Now consider $g_y(x)= x^ye^{-yx}=e^{y(\ln (x) -x)}$ with $y>0$ then $0\leq |g_y(x)| \leq |e^{-y} | \rightarrow 0$ as $y\rightarrow \infty$. This proves both pointwise and uniform convergence in $[0,\infty )$