Can someone explain to me this step from Harold Davenports 'Multiplicative Number Theory'?
\begin{equation} \psi(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^{s}} \end{equation} Since $\psi(s)$ is analytic for $\sigma> \frac{1}{2}$, it has expansion in powers of s-2 with radius of convergence at least $\frac{3}{2}$. This power series is \begin{equation} \psi(s) = \sum_{m=0}^{\infty} \frac{1}{m!}\psi^{(m)}(2)(s-2)^m \end{equation}
I assume we use Cauchy's Integral formula but I don't know if this is correct.
The function $\psi$ is an analytic function whose domain is $\Omega=\left\{z\in\mathbb C\,\middle|\,\operatorname{Re}z>\frac12\right\}$. For each $z_0\in\Omega$ there is a power series centered at $z_0$ which converges to $\psi(z)$ whose radius of convergence is at least the distance from $z_0$ to the boundary of $\Omega$. In the case of $2$, that distance is $\frac32$.
This is a general property of analytic functions: if the domain is $\Omega$ and if $z_0\in\Omega$, the radius of convergence of the Taylor series of the function centered at $z_0$ is at least the distance from $z_0$ to the boundary of $\Omega$.