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I am trying to complete exercise 10.9.f in Carothers Real Analysis. The ask is to provide a formal proof that $nxe^{-nx}$ converges pointwise and determine if it uniformly converges (if not, find a subinterval that does).

I propose the limit is 0, but I'm am not sure how to define the $n\geq N$ that will make $|nxe^{-nx}|<\epsilon$ true.

jmars
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1 Answers1

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The limit is $0$ for $x \geq 0$ but it is $-\infty$ for $x<0$. The convergence is not uniform on $(0,\infty)$ because $nxe^{-nx}=e^{-1}$ when $x=\frac 1 n$.

Hint for showing that the limit is $0$ for $x>0$: Use the inequality $e^{nx} >\frac {n^{2}x^{2}} 2$.

Kavi Rama Murthy
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  • I know this is a hint but I'm not sure where this inequality comes from and i'm more used to seeing these proofs structured in a way where one can explicitly require $n > \text{(some function of epsilon and maybe x)}$ – jmars Feb 24 '20 at 05:32
  • @jmars The inequality comes from Taylor expansion of exponential function. Using this show that $nxe^{-nx}=\frac {nx} {e^{nx}} \leq \frac {nx} {n^{2}x^{2} /2}$. Now you should be able to write down $N$ such that above quantity is lees than $\epsilon$ for $n >N$. – Kavi Rama Murthy Feb 24 '20 at 05:38
  • Oh interesting. I'm familiar with this Taylor expansion of e from my early calculus courses by Taylor expansions have not yet been treated in a rigorous way in Carothers/my analysis course. That is, I'm not sure if I'm allowed to invoke Taylor expansions for this proof. Is this the only way? – jmars Feb 24 '20 at 05:47
  • @jmars Have you studied L'Hopital's Rule? Using this rule it follows easily that $\frac t {e^{t}} \to 0$ as $t \to \infty$. Just putting $t=nx$ in this shows that the pointwise limit of your sequence is $0$. – Kavi Rama Murthy Feb 24 '20 at 05:57
  • I mayyy be able to get away with that. Thank you! – jmars Feb 24 '20 at 06:25