Please help me in prove / decline the point convergence, uniform convergence and near uniform convergence (comapact uniform convergence) of

$\sum_{n=1}^{\infty} f_n$ where

$f_n : [0, +\infty) \rightarrow \mathbb{R}$,

$ f_n = x^2 e^{-nx}$ for $x\in[0, +\infty)$,

$x\in \mathbb{R}$ and $n \geq 1$

This is an example from a long list in my homework, I want to watch how to solve this type of problem on it, then I'll do rest.

Thanks in advance!

  • $\begingroup$ How do you define "near uniform convergence"? $\endgroup$ – Pedro Tamaroff Apr 3 '13 at 23:16
  • $\begingroup$ @PeterTamaroff uniform on all closed intervals, not necessary uniform on whole domain $\endgroup$ – Joggi Apr 3 '13 at 23:18
  • $\begingroup$ @Joggi I'd call that "compact uniform convergence". Also $[0,\infty]$ instead of $[0,\infty)$ could produce some problems in this exercise. $\endgroup$ – kahen Apr 3 '13 at 23:30
  • $\begingroup$ @kahen Fixed in the question, thanks! $\endgroup$ – Joggi Apr 3 '13 at 23:32
  • $\begingroup$ Why both answers was deleted? $\endgroup$ – Joggi Apr 3 '13 at 23:51

Your function convergences pointwise (check this!) over $x>0$ to $$\frac{{{x^2}}}{{{e^x} - 1}}$$ and since $$\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{{e^x} - 1}}=0$$ we can say it converges pointwise to it all over $x\geq 0$. We want to show that given $\epsilon >0$ there exists an $N$ such that for any $x\geq 0$ $$\left| {{x^2}\sum\limits_{k = 1}^n {{e^{ - kx}}} - \frac{{{x^2}}}{{{e^x} - 1}}} \right| < \epsilon$$

whenever $n\geq N$.

This is always true if $x=0$, so assume $x>0$. We can then write the above as

$$\frac{{{x^2}}}{{{e^x} - 1}}{e^{ - x\left( {n + 1} \right)}} <\epsilon $$

Now, use $$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^x} - 1}} = 0 \cr & \mathop {\lim }\limits_{n \to \infty } {e^{ - nx}} = 0 \cr} $$

and $${e^{ - x\left( {n + 1} \right)}} < {e^{ - xn}}$$ to conclude.

  • $\begingroup$ Thank You for hint :-) I want to know also, what is general rule for checking near uniform convergence of function series $\endgroup$ – Joggi Apr 3 '13 at 23:14
  • $\begingroup$ @Joggi I changed my answer completely. I hope you don't mind. $\endgroup$ – Pedro Tamaroff Apr 3 '13 at 23:44

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