Suppose $f$ is analytic on $|z|<r$ except for a simple pole at $z=1$, and assume $g(z)= f(z) -R/(z-1)$, for $z\not = 1$ with $|z|<r$. Since $z=1$ is removable singularity of $g$, $g$ can be extended so that the extended function has an analytic antiderivative on the disk $|z|<r$. $R$ is the residue of $f$ at $1$. Then we have $$f(z) = g(z) + \frac{R}{z-1} = \sum_{n=0}^\infty(a_n-R)z^n$$ for $|z| <1$, where $a_0, a_1,...$ are the Taylor coefficients of $g$ at the origin.
How do we calculate to get the summation on the right?
It follows from the fact that for $\lvert z\rvert < 1$, $$\frac{1}{1 - z} = \sum_{n = 0}^\infty z^n$$