# Weierstrass M-test, and $\sum_{n=1}^\infty e^{-nx}x^n$

How can I use the Weierstrass M-test to show uniform convergence of $\sum_{n=1}^\infty e^{-nx}x^n$ on $[0,\infty )$?

I can't find any bounding sequences. I've tried to analyse the convergence on a smaller interval, but this didn't turn out too.

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Did you try to find where $e^{-nx} x^n$ is maximal? –  Jonas Teuwen Aug 1 '11 at 23:06
This is just a geometric series with common ratio $xe^{-x}$. –  Srivatsan Aug 2 '11 at 3:07

Note that $e^{-nx}x^n=(xe^{-x})^n$. Denote $f(x)=xe^{-x}$. For nonegative $x$, $f(x)$ is nonnegative, and $f$ has a global maximum when $f'(x)=x(-e^{-x})+(1)e^{-x}=0$, or $x=1$. (You can see this graphically, but if you need to establish it rigorously you also need to look at $f(0)=0$, $\lim_{x\to\infty}f(x)=0$, and $f''(1)<0$ to formally demonstrate a global maximum.) Then, given $f(1)=e^{-1}$, we have

$$e^{-nx}x^n = (xe^{-x})^n \le e^{-n}.$$

Since $\sum_{n=1}^\infty e^{-n}$ converges, as $|e^{-1}|<1$ and it's a geometric series, the M test shows that the original series converges uniformly.

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There is some argument about whether we should completely solve a possible homework problem without first waiting for the OP to return and reply to the hint. –  GEdgar Aug 1 '11 at 23:22
@GEdgar: I see your point. I'll try to be more patient in the future. –  anon Aug 1 '11 at 23:24
Thank you very much anon! By reading your answer, I've learned, and it feels great. My question was not a homework, I'm preparing for my final, and I'm doing random question in our textbook. I can assert the piece of machinery anon gave, I didn't have. But, I understand GEdgar point. –  Nicolas Essis-Breton Aug 2 '11 at 0:26