Finding the natural boundary of $\sum_{n=0}^{\infty}\frac{z^n}{n^k}$ for $k \ge 2$

Question

Let $k\ge 2$, given $$f(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^k}$$ It is easy to see that it converges for $|z|\le 1$, but how can it be analytically continued beyond the unit circle?

Hadamard proved that the radius of convergence of a power series in $\mathbb{C}$ extends to the nearest singularity of the function, so $f(z)$ must have a singularity on the boundary of the unit circle, which is why $f(z)$ cannot be extended to a full neighbourhood of the origin, as that would include the singularity. However, it may be extensible in some other directions.

For $k\ge2$, $f^{(k-1)}(z)=\sum_{n=k-1}^{\infty}\frac{n(n-1)\cdots (n-k+2)z^{n-k+1}}{n^k}$ blows up as you approach $z=1$, and this indicates that $z=1$ is a singularity of $f(z)$. But I cannot determine other singularity of $f(z)$ in this way.

Are other points on the boundary of the unit circle all singularity of $f(z)$?

Are there any other methods that would help determine the natural boundary of $f(z)$?

Any idea or hint would be appreciated!

Answer 1

Perhaps do it in stages. For $k \in \mathbb N$ define $$ f_k(z) = \sum_{n=1}^{\infty}\frac{z^n}{n^k} . $$ Now $$ f_0(z) = \sum_{n=1}^\infty z^n = \frac{z}{1-z} $$ has has finite radial limits everywhere on the bundary of the unit disk except at $z=1$.

Fix $k \ge 1$. Assume $f_{k-1}(z)$ has finite radial limits everywhere on the boundary except at $z=1$. Now $$ f_k'(z) = \sum_{n=1}^\infty\frac{z^{n-1}}{n^{k-1}} = \frac{f_{k-1}(z)}{z} $$ is analytic in the open unit disk (which is simply connected), so we may integrate to obtain $f_k$. This integral also has finite radial limits everywhere on the boundary except at $z=1$.

Answer 2

The main theorem in this 2016 paper states that a sequence $(c_k)_{k\ge0}$ is completely monotone if and only if the generating function $$F(z)=\sum_{k=0}^\infty c_k z^k$$ is (i.e., extends to be) a Pick function that is analytic and nonnegative on $(-\infty,1)$. A function is Pick if it is analytic on the upper half of the complex plane and has positive imaginary part there. W. F. Donoghue's book Monotone Matrix Functions and Analytic Continuation (Springer) covers the basic theory of Pick functions. Such functions are also called Herglotz or Nevalinna functions by others.

Since the function $g(x)=x^{-k}$ is completely monotone (having derivatives that alternate in sign), the sequence $c_k = n^{-k}$ is certainly completely monotone (which means that the forward differences of all orders alternate in sign).

By consequence, the function $$f(z)= -1 + \sum_{n=0}^\infty \frac{z^k}{n^k}$$ extends at least to the entire complex plane except possibly for the interval $[1,\infty)$ on the real axis. The fact that the series has radius of convergence $1$ and converges at $z=1$ indicates that $1$ is a singularity but not a pole. I suspect that a Pick function cannot have an essential singularity (though I could be wrong about that), so the singularity at $z=1$ should not be isolated. The location and nature of any other singularities in $[1,\infty)$ may depend on $k$. It would be interesting to be able to say more, but that's what I have for now.