Can a double series of $\ln(r)^m r^k$ converge to $r^\alpha$? Consider a series like
$$\sum_{k=0}^\infty\sum_{m=0}^\infty c_{k,m}\ln(r)^m r^k.$$
Can there be such a non-integer $\alpha$ and a set of $c_{k,m}$ that this series would converge to $r^\alpha$ in some neighborhood of $r=0$, and that partial sums would approximate $r^\alpha$ the better the closer $r$ is to $0$?

Both $\ln(r)$ and $r^\alpha$ are multi-valued functions in the complex plane, but I'm assuming we're taking compatible branches so that
$r^\alpha = \exp(\alpha \ln(r))$ for $r \ne 0$.  $r=0$ is a problem: see below.
If you didn't want to include $r=0$, just take $c_{km} = 0$ for $k > 0$ and 
$c_{0,m} = \alpha^m/m!$.  Thus your sum is
$$ \sum_{m=0}^\infty \frac{\alpha^m \ln(r)^m}{m!} = \exp(\alpha \ln(r)) = r^\alpha $$
Of course this won't work at $r=0$.  But you'd have that problem in any case: your expression is undefined at $r=0$.  However, if you use the convention $\ln(r)^m r^k = 0$ for $r=0$ when $k > 0$, you could take
$c_{1,m} = (\alpha-1)^m/m!$, $c_{k,m} = 0$ otherwise, and your sum is (for $r \ne 0$)
$$\sum_{m=0}^\infty \frac{(\alpha-1)^m \ln(r)^m r}{m!} = r \exp((\alpha-1)\ln(r)) = r^\alpha$$
while for $r=0$ both sides are $0$ if $\alpha > 0$.
