prove that the following series defines differentiable function on $(0,\infty)$

$$\sum_{n=0}^\infty x^2e^{-nx}$$ I tried to show:

1) $f_n(x)=x^2e^{-nx}$ is differentiable -simple.

2) $\sum_{n=0}^\infty x^2e^{-nx}$ convergent for one point- also simple (for instance $x=1$).

3)$\sum_{n=0}^\infty (f_n(x))'$ uniformly convergent.

I got stuck on 3). I'm was thinking about using M-test, but failed to find a series which is bigger than then this. I tried to find maximum to $(f_n(x))'$ and failed as well.

  • $\begingroup$ What do you mean by $\displaystyle \sum_{n = x}^\infty$? do you really want $\displaystyle \sum_{n = 1}^\infty$? or something like that? $\endgroup$ Oct 1, 2017 at 17:48
  • $\begingroup$ yes, sorry. one mistake many copies :) $\endgroup$
    – Mr.O
    Oct 1, 2017 at 17:51
  • $\begingroup$ use that $$\sum_{n=x}^\infty x^2e^{-nx}=\frac{x^2e^{x-x^2}}{e^x-1}$$ $\endgroup$ Oct 1, 2017 at 17:51
  • $\begingroup$ @Dr.SonnhardGraubner I need to show uniformly convergent for $\sum_{n=0}^\infty f_n(x)$ to use this sum? $\endgroup$
    – Mr.O
    Oct 1, 2017 at 18:04

1 Answer 1



You prove it is differentiable at $[a,+\infty) $ for arbitrary $a>0$.

then it will be differentiable at $(0,+\infty) $.

$$f'_n (x) =x (2-nx)e^{-nx} $$

the max is attained at $x=a$ since for large enough $n $, we have $$0 <2/n <a .$$

  • $\begingroup$ why differentiable $[a, \infty)$ implements to open interval $(0, \infty)$? $\endgroup$
    – Mr.O
    Oct 1, 2017 at 18:02
  • $\begingroup$ @Y.ofir For each $x>0$, there exists $a $ such that $0 <a <x $. $\endgroup$ Oct 1, 2017 at 18:09

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