Pointwise and uniform convergence

**thaopanda** · November 15th 2009, 11:13 AM

a) For each n $\in$ N consider $F_{n}$ : $[0, \infty) \rightarrow$ R be given by $F_{n}(x) := (x^n)e^{-nx}$ . Show that { $F_{n}$ } $_{n \in N}$ converges in a pointwise fashion to the zero function on $[0, \infty)$ .

b) Decide whether the convergence in part a is uniform on $[0, \infty)$ .

please help

**redsoxfan325** · November 15th 2009, 11:26 AM

Originally Posted by thaopanda

a) For each n $\in$ N consider $F_{n}$ : $[0, \infty) \rightarrow$ R be given by $F_{n}(x) := (x^n)e^{-nx}$ . Show that { $F_{n}$ } $_{n \in N}$ converges in a pointwise fashion to the zero function on $[0, \infty)$ .

b) Decide whether the convergence in part a is uniform on $[0, \infty)$ .

please help

a) Well, for all $x$ , $\lim_{n\to\infty}\frac{x^n}{e^{nx}}=0$ . One way to see this is that if you use L'Hopital's Rule $n$ times, you'll get $\lim_{n\to\infty}\frac{n!}{n!e^{nx}}=0$

b) The derivative of $x^ne^{-nx}$ is $nx^{n-1}e^{-nx}(1-x)$ , so the max on $[0,\infty)$ occurs at $x=1$ (for all $n$ ), and it follows that $x^ne^{-nx}\leq e^{-n}$ , which can clearly be made less than any $\epsilon$ . (Choosing $n>-\ln\epsilon$ does it.)

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