Taking $z = e^{-s}$, $U(z) = \sum_{t=0}^\infty f(t) z^t$ is a Maclaurin series with coefficients $f(t)$. If this converges for some $z \ne 0$, $U(z)$ is analytic in some disk $\{z: |z| < r\}$, i.e. $F(s) = U(e^{-s})$ is analytic for $\text{Re}(s) > \ln(1/r)$, periodic in $s$ with period $2 \pi i$, and $\lim_{\text{Re}(s) \to \infty} F(s) = f(0)$ exists. Moreover, the coefficients $f(t)$ can be extracted using the extended Cauchy integral formula: if $C$ is the positively-oriented circle $|z| = \rho$ with $0 < \rho < r$, corresponding to $\Gamma$ the line segment from $\ln(1/\rho)$ to $\ln(1/\rho) + 2 \pi i$, $ f(t) = \frac{1}{2\pi i} \oint_{\Gamma} \frac{U(z)}{z^{t+1}} \ dz = \frac{i}{2\pi} \int_C F(s) e^{st}\ ds$