Simple question about Taylor expansion of Dirichlet's eta function validity Consider the Taylor expansion of Dirchlet $\eta(s)$ at the point $s_0= \frac{1}{2} + it_0$  when using the classical representation of eta as $$\eta(s)= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{s}},$$ what is the disk for which the Taylor expansion of the former expression would be valid?  
I don't want to use any functional definition of $\eta$ or any relation to $\zeta$, just expanding the former definition. 
My understanding is since this expression of $\eta(s)$ converges for $\Re(s)>0$ and this is the expression used on the Taylor expansion, the result will be valid on the open disk centered at $t_0$ and with radius $\frac{1}{2}$, because bigger radios would traverse the area of convergence of the given serie.
Is this correct? Am I missing anything else ?
 A: Since $\eta$ is an entire function, the Taylor series about $s_0$ converges (to $\eta(s)$) on the whole plane.
Generally, if $U \subset \mathbb{C}$ is open, and $f$ holomorphic on $U$, the Taylor series of $f$ about $z_0 \in U$ converges (to $f$) on the every disk centred at $z_0$ that is contained in $U$. For $U = \mathbb{C}$, that means the Taylor series of $f$ about any point converges everywhere. For $U \subsetneqq \mathbb{C}$, there is a largest disk $D_1$ centred at $z_0$ that is contained in $U$. It can be that the (open) disk of convergence $D_2$ of the Taylor series is strictly larger than $D_1$. Then the Taylor series gives an analytic continuation of $f$ across part of the boundary of $U$. Note that in this case the Taylor series need not represent $f$ on all of $U \cap D_2$, this is only guaranteed to hold on the connected component of $z_0$ in $U \cap D_2$. A simple example of this phenomenon is given by the logarithm. Let $U = \mathbb{C}\setminus \{ t \in \mathbb{R} : t \leqslant 0\}$, and $f$ the principal branch of the logarithm. The Taylor series of $f$ about $z_0 = -2 + i$ has radius of convergence $\sqrt{5}$ (the distance from $z_0$ to the branch point $0$), but it only represents $f$ on the part of that disk contained in the upper half-plane. On $\{ z : \lvert z - z_0\rvert < \sqrt{5}, \operatorname{Im} z < 0\}$ the Taylor series converges to $f(z) + 2\pi i$.
