I am nearly complete in my understanding of this beautiful theorem of Dirichlet. On page 34 of Davenports book (Multiplicative number theory) the following breakdown is present and I do not understand where the expansion of powers comes in
\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} The value for $\psi^{(m)}(2)$ can be obtained from the Dirichlet series in Equation (29) from and we get: \begin{equation} \psi^{(m)}(2) = (-1)^m \sum_{n=1}^{\infty} a_n (\text{log}n)^m n^{-2}=(-1)^m b_m \end{equation} where $b_m > 0$. So we have \begin{equation} \psi(s) = \sum_{m=0}^{\infty} \frac{1}{m!} b_m (2-s)^m \end{equation} and this holds for $|s-2|<\frac{3}{2}$. If $\frac{1}{2}<s<2$ and since all terms are non negative we have $$\psi(s) \geq \psi(2) \geq 1$$ and this contradicts the fact that $\psi(s) \to 0$ ````
Does the second line have something to do with Cauchy's integral formula? Can someone explain to me the 2nd step, and also the very last line?