# Properties of analytic continuations of zeta functions

Let $$\{u_k\}_{k\in\mathbb{N}}$$ be an increasing (or even nondecreasing?) sequence of positive reals, limiting to infinity.

For the series $$f(s) := \sum_{k=1}^{\infty} (u_k)^{-s}$$ with $$s$$ a complex variable, by the results stated in the comments and answers here, we know there exists $$\sigma \in [-\infty,+\infty]$$ such that the convegence set in $$\mathbb{C}$$ of $$f(s)$$ is $$\{\mathrm{Re}(s) > \sigma\}$$.

I am also wondering about the following:

1. Is $$f(s)$$ holomorphic with a unique continuation to a meromorphic function on $$\mathbb{C}$$?

If yes, then:

2. If $$\sigma\notin\{\pm\infty\}$$, does the continuation of $$f(s)$$ have a single pole located exactly at $$\sigma +0i \in \mathbb{C}$$?

For example, for $$u_k = k$$ we get $$f$$ as the Riemann zeta function, and the answers to the above questions are yes, with $$\sigma = 1$$.

For background, I am thinking about this in the context of the above series being the zeta function of an elliptic differential operator. After eqn (1.20) in this paper, the author claims that if $$u_k = \lambda_k$$ is the (ordered) spectrum of a positive elliptic differential operator in dimension $$n$$, then $$\sigma \leq n/2$$.

Edit: removed "analytic" from the continuation.

• Note that if $u_k=e^{p_k}$ where $p_k$ is the $k$ the prime the series (under the obvious substitution) becomes a power series with natural boundary unit circle, so as Dirichlet series the abscissa $\Re s =1$ is natural boundary Commented Jul 5, 2022 at 0:12

In general, Dirichlet series do not have meromorphic (much less analytic) continuations to $$\mathbb C$$ (even ignoring the common pole(s) at $$s=1$$).
Googling "Estermann phenomenon", you will find out that Estermann showed in 1928 that many natural-looking Dirichlet series, even with Euler products, have demonstrable natural boundaries. E.g., $$\sum_n d(n)^3/n^s$$, where $$d(n)$$ is the number of positive divisors of $$n$$.
• Thank you for this example and explanation! However, I think the example series you are using is not of the type I described in my question. (Since $d(n)^3 n^{-s}$ can't be written as $u_n^{-s}$, for arbitrary $s$, for some $u_n$ depending only on $n$.) Commented Jul 4, 2022 at 19:00