Suppose you have $$f_n: \mathbb{R} \rightarrow \mathbb{R} $$ $$x \mapsto \frac{x^n e^{-x}}{n!} $$

I am asked to find the limit of this function and prove its uniform convergence.

For finding the limit(basically showing that it converges pointwise) I did this:

Using the stirling's formula, I found that $$f_n(x)\sim \frac{x^n e^{-x}}{n^n e^{}-x \sqrt{2 \pi n}} = \left(\frac{x}{n}\right)^n \frac{e^{-x}}{e^{-n}} \frac{1}{\sqrt{2 \pi n}}=\left (\frac{x}{n}\right)^n e^{n-x} \frac{1}{\sqrt{2 \pi n}} = \left(\frac{xe^{1- \frac{x}{n}}}{n}\right)^n \frac{1}{\sqrt{2 \pi n}}$$ Thus it converges to the null function.

For the uniform convergence, I derived the function and searched when it reached its maximum.

The derivative of $f_n(x)$ is $$f'_n(x)=\frac{1}{n!}\left(\frac{nx^{n-1}-x^n}{e^x}\right)$$ Thus is equals to zero whenever $x=n$. This allows me to state that the function $f_n$ reaches it's maximum whenever $x=n$, yet $f_n(n) \sim \frac{1}{\sqrt{2 \pi n}}$

Now I am asked to find $$ \underset{n \rightarrow \infty}{\lim} \int_{0}^{\infty}f_n(t)dt $$, but thinking logically, I am most likely asked to integrated the limit function because I proved the uniform convergence. Yet I don't know exactly to what it converges. Can someone help me out?

  • $\begingroup$ Every $f_n$ is a non-negative function with unit integral over $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Feb 22 '17 at 13:03

I think the goal of the exercise was to give an example of sequence of function which converge pointwisely and uniformly to zero, yet their intergrals are all equal to $1$.

You have $$\int_0^\infty x^{s-1}e^{-s}ds = \Gamma (s),$$ where $\Gamma$ is the Gamma-function. For natural $n$ you can find - by integration by parts - that $\Gamma(n+1)=n!$, hence for your problem $$\forall n\in \Bbb N \quad\int_0^{\infty}\frac{x^ne^{-x}}{n!}dx = 1.$$

Digression. We know that if $f_n\to f$ uniformly on a set $\Omega$, with $f_n$ and $f$ being in $L^1(\Omega)$ and $|\Omega| <\infty$, then $f_n\to f$ in $L^1(\Omega)$. In your exercise, however, we do not have $|\Omega|<\infty$, hence the convergence in $L^1$ is not guaranteed and we build an explicit counterexample.


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