= Isolating a logarithmic singularity to test a power-series endpoint
{title2=$f(z)=A\log(1-z/R)+H(z)$}
Suppose $A\ne0$ 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 $-A/(nR^n)$ 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 $R=\log2$ 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 $R\pm2\pi i$, so a <holomorphic logarithm> exists on a larger disk. This shows why the <Taylor series> converges at $x=-\log2$ even though its singular positive endpoint is excluded.
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