Isolating a logarithmic singularity to test a power-series endpoint ($f(z)=A\log (1-z/R)+H(z)$)

Suppose $A\ne 0$ and $R>0$, and $H$ is an analytic function on a disk of radius strictly larger than $R$. The Taylor series at zero of $f(z)=A\log (1-z/R)+H(z)$ has radius of convergence $R$. Its coefficients are the sum of $-A/(n{R}^n)$ and the more rapidly decreasing coefficients of $H$. At $x=-R$, the logarithmic part is an alternating harmonic series and converges conditionally; at $x=R$ it diverges. The power series for $H$ converges absolutely at both endpoints and therefore cannot change either conclusion. The real convergence interval is $[-R,R)$.

For $f(z)=\log (2-e^z)$, take $R=\log 2$ and $H(z)=\log [(2-e^z)/(1-z/R)]$, with its branch normalized by $H(0)=0$. The quotient has nonzero removable value $2R$ at $z=R$. Its next zeros are at $R\pm 2\pi i$, so a holomorphic logarithm exists on a larger disk. This shows why the Taylor series converges at $x=-\log 2$ even though its singular positive endpoint is excluded.