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. $\blacksquare$
Your help is greatly appreciated!
Best Answer
Proposition. The following approximation holds true: $$f(s)=\frac{1}{2}\log s+O(1),\qquad s\in(0,1).$$ Proof. Following Carlo Benakker, we remark that $$\zeta(s)+\frac{s}{1-s}=-\frac{1}{2}+O(s),\qquad |s|<2.$$ Indeed, the left-hand side is an entire function with value $-1/2$ at $s=0$. Using this approximation, we obtain for $s\in(0,1)$ that $$\begin{align*} f_1(s)&=\sum_{n=1}^{\lceil 1/s\rceil} \left (-\frac{1}{2n}+O(s)\right )\\ &=-\frac{1}{2}\sum_{n=1}^{\lceil 1/s\rceil}\frac{1}{n}+O(1)\\ &=\frac{1}{2}\log s+O(1). \end{align*}$$ It remains to show that $f_2(s)$ is bounded for $s\in(0,1)$. To see this, we use the integral representation $$\zeta(s)+\frac{s}{1-s}=-s\int_1^\infty\{u\}u^{-s-1}du,\qquad \Re s>0,$$ which yields that $$f_2(s)=-s\sum_{n=\lceil 1/s\rceil+1}^{\infty} \int_1^\infty\{u\}u^{-ns-1}du,\qquad \Re s>0.$$ By Fubini's theorem, we can interchange the summation and the integration, so that $$f_2(s)=-s\int_1^\infty\frac{\{u\}}{u^{1+s\lceil 1/s\rceil}(u^s-1)}\,du,\qquad \Re s>0.$$ Assume now that $s\in(0,1)$. Then $s\lceil 1/s\rceil \geq 1$, and $u^s-1>s\log u$ for all $u>1$, whence $$|f_2(s)|\leq \int_1^\infty\frac{\{u\}}{u^{2}\log u}\,du,\qquad s\in(0,1).$$ The last integral converges, and it is independent of $s$, hence we are done.
In fact, this upper bound is optimal:
Added. For $s\to 0+$, we have that $s\lceil 1/s \rceil \to 1$, and $(u^s-1)/s\to\log u$ for all $u>1$, hence Lebesgue's dominated convergence theorem yields $$\lim_{s\to 0+}f_2(s)=-\int_1^\infty\frac{\{u\}}{u^2\log u}\,du.$$