If I apply the root test to the series of functions $$\sum_{k=1}^\infty \ln(k)x^k$$
I end up with the limit $$x\cdot\lim_{k\to\infty}\sqrt[k]{\ln(k)}=x\cdot\infty>1$$
which shows this series does not even converge pointwise. I realize this argument is quite weak, and probably wrong. But I'm running out of ideas at this point. The big problem here, is that the limit function seems inaccessible, so we cannot apply many of the theorems. If I had the limit function, I could examine its continuity on the interval $[0,1)$. If it was not continuous, then this series would not be uniformly convergent. What I am quite sure about, is that this is series is not uniformly convergent.
Best Answer
You don't always need to know the sum to examine uniform convergence of a series.
A necessary and sufficient condition for uniform convergence is the uniform Cauchy criterion. This states that the convergence of a series $\sum f_k(x)$ is uniform for $x \in D$ if and only if for all $\epsilon > 0$ there exists $N(\epsilon) \in \mathbb{N}$ such that for all $m \geqslant n \geqslant N(\epsilon)$ and all $x \in D$, we have
$$\tag{*}\left| \sum_{k=n}^m f_k(x)\right|< \epsilon$$
Taking $m = n$ we seee that for uniform convergence of the series it is necessary that $|f_n(x)| < \epsilon$ for all $n \geqslant N(\epsilon)$ and $x \in D$. This is equivalent to the requirement that
$$\lim_{n \to \infty}\sup_{x \in D}f_n(x) = 0$$
With this result we see that the series in question fails to converge uniformly for $x \in [0,1)$ since
$$\lim_{n \to \infty} \sup_{x \in [0,1)} \ln n \, x^n= \lim_{n \to \infty}\ln n = +\infty $$