# On the asymptotic behaviour of the series $\sum_{n=1}^{\infty} \left ( \frac{\zeta(ns)}{n}+\frac{s}{1-ns}\right )$ near $s=0$

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 $$0. We can then write $$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 $$f_2(x)=\mathcal{O}(1)$$.

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.

• Why (or in what sense) do you think the infinite sum on $n$ converges? As you observe, each term looks like $-1/2n +O(s)$. Commented Jan 23 at 16:34
• It, of course, diverges at $s=0$. Commented Jan 23 at 16:35
• Why should it converge anywhere near $s=0$? Commented Jan 23 at 16:36
• Yes, the proof please. Commented Jan 23 at 21:20
• In my post below, I confirm the claimed approximation for $f(s)$. Commented Jan 24 at 13:28

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 $$\lim_{s\to 0+}f_2(s)=-\int_1^\infty\frac{\{u\}}{u^2\log u}\,du.$$

• Wonderful argument! So, now we know that $f_1(s)=\frac{1}{2} \log s + \mathcal{O}(1)$ and that $f_2(s)=\mathcal{O}(1)$. But, can we actually evaluate $C=\lim_{s \rightarrow 0} f_2(s)$? And, if so, can we claim right away that $f_2(s)=C+\mathcal{O}(s)$?. That would be of great help to me! Commented Jan 24 at 13:33
• @TianVlašić See my "Added" section. Commented Jan 24 at 13:41
• @TianVlašić We cannot claim right away that $f_2(s)=C+O(s)$. I gave you the value of $C$. Now you can write $f_2(s)-C$ as a single integral over the line $u>1$, and analyze that integral similarly as I analyzed the original integral. I am lazy to work this out, but I am confident that $f_2(s)=C+O(s)$ will follow. I guess you will need to use the second order Taylor expansion of $u^s-1$ around $s=0$. Commented Jan 24 at 13:53
• I think this is enough for me to work the rest out by myself. Thank you once again! Commented Jan 24 at 13:56
• $\int_1^\infty\frac{\{u\}}{u^2\log u}du=0.75036751215075390992131242125469306130...$ Commented Jan 28 at 18:41

We seek the small-$$s$$ asymptotics of the sum $$f(s)=s\sum_{n=1}^\infty g(ns),\;\;g(x)=\frac{\zeta(x)}{x}+\frac{1}{1-x}.$$ First note that $$\lim_{s\rightarrow 0}\sum_{n=\lfloor{2/s}\rfloor+1}^\infty sg(ns)=\int_2^\infty g(x)\,dx=-0.48,$$ hence the summation range $$n\gtrsim 2/s$$ contributes an amount of order $${\cal O}(1)$$ to $$f(s)$$.
For $$n\lesssim 2/s$$ you can expand $$\frac{\zeta(ns)}{n}+\frac{s}{1-ns}=\frac{-1}{2 n}+{\cal O}(s),$$ and sum $$\sum_{n=1}^{\lfloor 2/s\rfloor}\frac{-1}{2 n}=\tfrac{1}{2}\ln s +{\cal O}(1),$$ resulting in $$f(s)=\tfrac{1}{2}\ln s+{\cal O}(1).$$ A numerical evaluation confirms this is the correct asymptotics:

Blue is the numerically evaluated sum, as a function of real $$s$$, red is $$\tfrac{1}{2}\ln s-1$$ on a log-linear scale; the slopes agree.

For complex $$s$$ (with $$\operatorname{Re}s>0$$) the absolute value governs the asymptotics, $$|f(s)|\rightarrow \tfrac{1}{2}\ln|1/s|+{\cal O}(1)$$. The plot below shows $$|f(s e^{i\pi/4})|$$ (blue) and $$1+\tfrac{1}{2}\ln |1/s|$$ (red), again with good correspondence.

• To make this formal, I suggest, as a starting point, to consider the sum $\sum_{n=1}^{\left \lfloor \frac{1}{s} \right \rfloor}\frac{-1}{2n}$. Commented Jan 23 at 21:58
• I added the case of complex $s$, with the $\tfrac{1}{2}\ln|1/s|$ asyptotics. Commented Jan 23 at 23:15
• That's great, but I think it is much wiser to focus on the restriction $s>0$. Commented Jan 23 at 23:25
• I have added the evaluation of the large-$n$ range, where the sum can be replaced by the integral to leading order in $s$. Commented Jan 24 at 7:47
• How do we know the relation $\lim_{s \rightarrow 0}\sum_{n=\lfloor \frac{2}{s} \rfloor +1}^{\infty} s g(ns)=\int_{2}^{\infty} g(t) dt$? Commented Jan 24 at 9:14