Since the radius of convergence $R$ for the Taylor series of $\ln(z)$ around $1$ is $1$, i.e.
$$ R\left\{\ln\left(1+z\right) = \sum \frac{(-1)^{k-1}}{k}x^k \right\} = 1 ,$$
does this mean that for
$$\ln\left(a+x\right) = \ln\left(a\left(1+\frac{x}{a}\right)\right) = \ln\left(a\right) + \ln\left(1+\frac{x}{a}\right)$$
the Taylor series converges for
$$-1 < \frac{x}{a} < 1$$
and hence for
$$-a < x < a ?$$
If so, this would seem to imply that if I take "large enough" $a$, then I can get convergent Taylor series expansion for $\ln(x)$ on
$$(0,2a)$$
covering a large range of positive reals?
If so, then if I needed to get convergence on $(b,c)$, is there a "good way" to choose $a$ so that the series converges fast at both ends"? (I suspect that the rate of converges for $x<a$ is different from that for $x>a$, so that choosing $a$ not simply as a mid point of $(0,c)$ might be better.)
You are correct. You can also use the fact that$$\log(z+a)=\log(a)+\sum_{n=1}^\infty(-1)^{n-1}\frac{z^n}{na^n}$$and that, assuming that $a>0$, the radius of convergence of this power series is (again) $a$.