$$\sum_{n=1}^{\infty}\frac{n^n}{n!x^n},x>0$$ Using a change of variables and applying the usual power series theorems I've found out that the series converges pointwise and absolutely for $x>e$, and converges uniformly in $[e+\delta,\infty[,\delta>0$. I don't know how to treat the case $]e,+\infty[$ for uniform convergence though. Anyone?
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Theorem: If a sequence $f_k$ of continuous functions converges uniformly on $]a,\infty[$, and all functions are defined and continuous on $[a,\infty[$, then the sequence converges uniformly on $[a,\infty[$.
For by continuity, we have
$$\sup_{x \in ]a,\infty[} \lvert f_k(x) - f_m(x)\rvert = \sup_{x\in [a,\infty[} \lvert f_k(x) - f_m(x)\rvert.$$
Apply the theorem to the sequence of partial sums, and deduce that if the convergence were uniform on $]e,\infty[$, then the series would in particular also converge for $x = e$.
Daniel Fischer
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The series doesn't converge uniformly on $(e,+\infty)$ since for $x=e+\frac1n$ we have using the Stirling's approximation
$$\frac{n^n}{n!x^n}=\frac{n^n}{n!\left(e+\frac1n\right)^n}\sim_\infty\frac{1}{\sqrt{2\pi n}\left(1+\frac1{en}\right)^n}\sim_\infty\frac{e^{-1/e}}{\sqrt{2\pi}}\frac1{\sqrt n}$$ and the series $$\sum_{n\ge1}\frac1{\sqrt n}$$ is divergent.