I am interested in determining the behaviour of the the series/function

$$f(s)=\sum_{n=1}^{\infty} \left ( \frac{\zeta(ns)}{n}+\frac{s}{1-ns}\right )$$

near $s=0$. It is clear that $f(0)$ is undefined. However, $f(s)$ is well-defined for all complex variables $s$ with $\mathrm{Re}(s)>0.$

All "classical" methods have failed me; Expanding $\zeta(ns)$ near $ns=0$ as a finite number of terms in the Laurent series for $\zeta$ results in an divergent overall sum. I have also tried writing the sum as a partial sum and taking the limit but to no avail.

**NOTE:** I was, at first, only looking for a hint. But since the question seems to be non-trivial, I would like to rigorously prove the expansion of the aforementioned series.

@Carlo Beenakker has in his answer to this post conjectured the expansion

$$f(x)=\frac{1}{2} \log x+\mathcal{O}(1).$$

Now let $x$ be a real variable with $0<x \leq 1$. We can then write $f(x)=f_1(x)+f_2(x),$ where

$$f_1(x)=\sum_{n=1}^{\lceil \frac{1}{x}\rceil} \left ( \frac{\zeta(nx)}{n}+\frac{x}{1-nx}\right )$$

and

$$f_2(s)=\sum_{n=\lceil \frac{1}{x}\rceil+1}^{\infty} \left ( \frac{\zeta(nx)}{n}+\frac{x}{1-nx}\right ).$$

Indeed, $f_1(x)$ does probably have the desired asymptotic expansion

$$f_1(x)=\frac{1}{2} \log x -\frac{\gamma}{2}+\mathcal{O}(x).$$

Thus, I think it is, after proving this, sufficient to prove that $f_2(x)=\mathcal{O}(1)$.

**Further tasks:**

1.) Proving the asymptotic expansion of $f_1(x)$.

2.) Explicitly evaluating the following limit:

$$\lim_{s \rightarrow 0} f_2(x)= \lim_{x \rightarrow 0} \sum_{n=\lceil \frac{1}{x}\rceil+1}^{\infty} \left ( \frac{\zeta(nx)}{n}+\frac{x}{1-nx}\right ).$$

**Appendix:**

On the request of @Stropple, I prove the following proposition:

**Proposition**. $f$ is a complex-valued function on the right half-plane $ \left \{s \in \mathbb{C} : \mathrm{Re}(s) > 0 \right \}.$

*Proof*. Let $s \in \mathbb{C}$ with $\mathrm{Re}(s)>0$. Assume $\mathrm{Re}(s) > 1$. Upon making use of the inequality

$$\forall z \in \mathbb{C}, \ \mathrm{Re}(z)>1: \left | \frac{\zeta(z)}{z}+\frac{1}{1-z} \right | \leq \frac{1}{\mathrm{Re}(z)(\mathrm{Re}(z)-1)},$$

we have

$$\left |f(s) \right | = \left |\sum_{n=1}^{\infty} s \left ( \frac{\zeta(ns)}{ns}+\frac{1}{1-ns} \right) \right | \leq \left | s \right | \sum_{n=1}^{\infty} \left |\frac{\zeta(ns)}{ns}+\frac{1}{1-ns} \right | \leq \left | s \right | \sum_{n=1}^{\infty} \frac{1}{\mathrm{Re}(ns)(\mathrm{Re}(ns)-1)}=-\frac{\left |s \right| \left(\gamma+\psi \left ( 1-\frac{1}{\mathrm{Re}(s)} \right ) \right)}{\mathrm{Re}(s)}$$

where $\psi$ is the digamma function. By the choice of $s$, the resulting upper bound is finite.

Assume now $0 < \mathrm{Re}(s) \leq 1$. Then, we can write

$$\left |f(s) \right| = \left |f_1(\mathrm{Re}(s))+f_2(\mathrm{Re}(s)) \right| \leq \left |f_1(\mathrm{Re}(s)) \right|+\left |f_2(\mathrm{Re}(s)) \right|$$

If $\mathrm{Re}(s) \notin \left \{\frac{1}{k} : k \in \mathbb{N} \right \}$, it is clear that $\left | f_1(\mathrm{Re}(s)) \right|$ is finite. However, even when $\mathrm{Re}(s)=\frac{1}{k}$ for some $k \in \mathbb{N}$, $\left | f_1(\mathrm{Re}(s)) \right|$ is finite (the term $n=k$ in the definition of $f_1$ is taken in the sense of a limit). It is thus sufficient to bound $\left |f_2(\mathrm{Re}(s)) \right|$. By a similar argument, we have

$$\left |f_2(\mathrm{Re}(s)) \right|=\left |s \sum_{n=\lceil \frac{1}{\mathrm{Re}(s)}\rceil+1}^{\infty} \left ( \frac{\zeta(ns)}{ns}+\frac{1}{1-ns}\right ) \right| \leq \left | s \right| \sum_{n=\lceil \frac{1}{\mathrm{Re}(s)}\rceil+1}^{\infty} \left | \frac{\zeta(ns)}{ns}+\frac{1}{1-ns}\right| \leq \left | s \right| \sum_{n=\lceil \frac{1}{\mathrm{Re}(s)}\rceil+1}^{\infty} \frac{1}{\mathrm{Re}(ns)(\mathrm{Re}(ns)-1)} \leq \frac{\left | s \right| \left (\psi \left ( \left \lceil \frac{1}{\mathrm{Re}(s)} \right \rceil +1 \right )-\psi \left (\lceil \frac{1}{\mathrm{Re}(s)}\rceil+1-\frac{1}{\mathrm{Re}(s)} \right ) \right )}{\mathrm{Re}(s)}.$$

By the choice of $s$, the resulting upper bound is finite.

Your help is greatly appreciated!

4more comments