$\displaystyle
f(z)=\sum_{n=0}^{\infty}(2+z)^n, \ \ (|z+2|<1)
$
$\displaystyle
g(z)=-2 \sum_{n=1}^{\infty}\left( \frac{1}{3+2x}\right)^n, \ \ (|z+3/2|>1/2)
$
The series defining function $\displaystyle f(z)$ is a geometric series which sums to:
$\displaystyle
f(z)=\frac{-1}{1+z}, \ \ (|z+2|<1)
$
Similarly the series defining $\displaystyle g(z)$ may also be summed with the
aid of the geometric series formula to:
$\displaystyle
g(z)=2-2 \sum_{n=0}^{\infty}\left( \frac{1}{3+2x}\right)^n
$$\displaystyle =2-\frac{2}{1-\frac{1}{3+2z}}=\frac{-1}{1+z} , \ \ (|z+3/2|>1/2)
$
Thus $\displaystyle f(z)$ and $\displaystyle g(z)$ are identical on the overlap of their regions of convergence, and so are analytic continuations of one another.
RonL