# 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.