Is $\sum_{n=1}^{\infty} {x^2 e^{-nx}}$ uniformly convergent in $[0,\infty)$?

So I started by saying that by the geometric series test where $a=x^2$ and $|r| = |\frac{1}{e^x}| \leq 1$, the series converges pointwise.

But how do I exactly prove that it converges uniformly? I am quite sure it is by weistrass test but I can not find an upper bound to compare it to! Any direction would be appreciated!

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    $\begingroup$ Have you tried the M-test? $\endgroup$ – oldrinb May 28 '13 at 20:15
  • $\begingroup$ As I said in my original post that I was sure it was by the weistrass $M$ test. but all I get is that $\frac{x^2}{e^{nx}} \leq \frac{x^2}{e^x}$, but that almost gives me nothing (I think, at least.) $\endgroup$ – TheNotMe May 28 '13 at 20:16

Let $$f_n(x)=x^2e^{-nx}$$ then we have $$f'_n(x)=e^{-nx}\left(2x-nx^2\right)=0\iff x=0\ \text{or}\ x=\frac{2}{n}$$ so $$||f_n||_\infty=f_n\left(\frac{2}{n}\right)=\frac{4}{n^2}e^{-2}$$ hence the series $\displaystyle \sum_{n=1}^\infty ||f_n||_\infty$ is convergent and then the series $\displaystyle \sum_{n=1}^\infty f_n$ is uniformly convergent.

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  • $\begingroup$ Clear and easily understandable. Thank you so much. $\endgroup$ – TheNotMe May 28 '13 at 20:19
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 May 28 '13 at 20:20
  • $\begingroup$ @SamiBenRomdhane, You used the M-test, right? $\endgroup$ – 17SI.34SA Jun 23 '13 at 22:12

You can compute the remainder term explicitly, using the formula for the geometric series: $$ r_N(x) = \sum_{n=N}^\infty x^2 e^{-nx} = x^2e^{-Nx} \sum_{j = 0}^\infty e^{-jx} = \dots $$ Now find the maximum of $r_N$ on $[0, \infty)$. (It's obviously a positive function.)

If this maximum tends to $0$ as $N \to \infty$, then the convergence is uniform. Otherwise it isn't.

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