Given $m>n$, for $x\in[0,\infty)$, we have $ \Bigl|\,\sum_{k=n}^m (-1)^{k+1}{1\over k}e^{-kx}\,\Bigr| \le {1\over n}e^{-nx}\le {1\over n}. $
It follows that $\sum\limits_{n=1}^\infty (-1)^{n+1}{1\over n}e^{-nx}$ is uniformly Cauchy on $[0,\infty)$ and, thus, uniformly convergent on $[0,\infty)$.
Below, are sketched the first few partial sums $S_k=\sum\limits_{n=1}^k (-1)^{n+1}{1\over n} e^{-nx}$ of the series. Note how they "alternate":

More generally, if $(f_n)$ is a decreasing sequence of nonnegative functions that converge uniformly to $0$ on the set $I$, then the series $\sum\limits_{n=1}^\infty (-1)^n f_n$ is uniformly convergent on $I$.