Find the analytic continuation of $f(z)=\sum_{1}^{\infty} \frac{z^n}{n^2}$ Find the analytic continuation of $f(z)=\sum_{1}^{\infty} \frac{z^n}{n^2}$. This function clearly converges on the closed disk by the usual convergence test. To continue, my strategy is to find a closed form of $f$. One thing I tried was to differentiate it to see whether its derivative looks nice enough but it failed.
 A: Set
$$g(z) = \frac{-\log(1-z)}{z},$$
initially defined for $z\in \mathbb C \setminus ([1,\infty)\cup\{0\}).$ Here $\log$ denotes the principal value logarithm. If we define $g(0)=1,$ then the singularity at $0$ is removable. So we may regard $g$ as analytic on $U =\mathbb C \setminus [1,\infty).$
Now $U$ is simply connected, and therefore $g$ has an antiderivative $G$ in $U.$ Actually we don't need this result, as the definition
$$G(z) = \int_0^z g(w)\,dw$$
gives us an antiderivative of $g$ in $U$ directly. Here the contour from $0$ to $z$ is the line segment from $0$ to $z.$ You can verify that $G'(z)=g(z)$ in the usual way.
Now for $|z|<1,$ $f'(z) = g(z).$ It follows that for $|z|<1,$ $f(z)= G(z)+C.$ Since $f(0)=0=G(0),$ $C=0.$ So we have $f=G$ in the open unit disc, and since $G$ is analytic in $U,$ $f$ extends to be analytic in $U.$
So there you have it: $G$ is the analytic continuation of $f$ to all of $U.$
A: Sounds like under some assumptions,
$$
f'(z) = \sum_n z^{n-1}/n,
$$
so
$$
\frac{d}{dz} \left[zf'(z)\right] = \sum_n z^{n-1}
$$
