# prove that the following series ($\sum_{n=0}^\infty x^2e^{-nx}$) defines differentiable function on $(0,\infty)$

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.

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

hint

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 $$0 <2/n <a .$$

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