How to show that $\sum_{n=1}^{+\infty}\frac{z^{2n}}n$ can be analytically continued for $|\Re z|<1$? Let's consider the following power series: $\sum_{n=1}^{+\infty}\frac{z^{2n}}n$
I think that the convergence radius is $R=1$.
Then the function given by $f(z)=\sum_{n=1}^{+\infty}\frac{z^{2n}}n$ is holomorphic in the unit disk. To prove that it cannot be continued on $\mathbb C$, I want to use that the series diverges at $z=1$ but I don't know how to rigorously state that.
Finally, I am wondering how to show that it can be continued for $|\Re z|<1$ (if it can)?
 A: The series evaluates to $-\log(1-z^2)$. Using the principal $\log$ with branch cuts on the nonpositive real axis, it is easy to show that $1-z^2$ is a nonpositive real number iff $z$ is real and of magnitude at least $1$. Since $|\operatorname{Re}(z)|<1$ does not touch these cuts, it follows that there is an analytic function on $|\operatorname{Re}(z)|<1$ that agrees with the series on the latter's convergence disk of $|z|<1$.
There is no analytic continuation to $\mathbb C$ because $\lim_{z\to1^-}\sum_{n=1}^\infty\frac{z^{2n}}n=+\infty$, whereas the limit of an analytic function is finite in any direction and at any point.
A: Let $\operatorname{Log}\colon\Bbb C\setminus(-\infty,0]$ be the main branch of the logarithm. Then, for each $z\in D(0,1)$,$$-\operatorname{Log}(1-z^2)=\sum_{n=1}^\infty\frac{z^{2n}}n.$$But then$$\begin{array}{ccc}\{z\in\Bbb C\mid|\operatorname{Re}(z)|<1\}&\longrightarrow&\Bbb C\\z&\mapsto&-\operatorname{Log}(1-z^2)\end{array}$$is an analytic continuation of $\sum_{n=1}^\infty\frac{z^{2n}}n$.
