# Exploring the Dirichlet series of the sum of remainder function

I wolud like to learn and understand more some basic facts about Dirichlet series, for wich I want explore the following function, that is called the sum of remainders function, A004125 as Sloane's sequence. $$S(n)=\sum_{k=1}^n \text{nmod k}.$$

Question 1. Can you explore the abscise of absolute convergence $\sigma_a$ for the Dirichlet series $$\sum_{n=1}^\infty\frac{S(n)}{n^s}?$$

Facts that I thought are, the first that it is known that this function is related with the sum of divisor function $\sigma(m)=\sum_{d\mid m}m$, by $$\sigma(n)+S(n)=S(n-1)+2n-1,$$ which holds for each $n>1$. Thus I can think in the diference $|f(n)-f(n-1)|$ and perhaps use the triangle inequalities or the knowdlegde that we have about the sum of divisor function. Too if we multiply by $n^{-s}$ we can get the Dirichlet series for $\sigma>2$ for the sum of divisor function and $\zeta(s)$ and $\zeta(s-1)$, and perhaps from here you know how work with $\sum_{n=2}^\infty\frac{S(n)}{n^s}$ and $\sum_{n=1}^\infty\frac{S(n)}{(n+1)^s}$.

Following Apostol's book, Introduction to Analytic Number Theory (now page 226), if we can get a bound $|S(n)|$, I would like write the bound for the tails $|\sum_{n=N}^\infty\frac{S(n)}{n^s}|$ as is stated in Lemma 1, for a $\sigma\geq c\geq \sigma_a$, which is a direct application of such lemma, but I ask if one time that you know previous bound for $|S(n)|$, then

Question 2. Can you get an upper bound for $$|\sum_{n=N}^\infty\frac{S(n)}{n^s}|?$$ And, can you conjecture or deduce the abscise corresponding with a half-plane for which the Dirichlet series $\sum_{n=1}^\infty\frac{S(n)}{n^s}$ is never zero?

Thanks. I would like to ask more about these pages, for example the question concerning the limit uniformly to S(1)=0, but I assume that I am learning slowly.

Update: The remarks that were deduced in comments:

If we're only looking at $k \geq n/2$, then $n \bmod k = (n-k)$, and summing that gives approximately $n^2/8$.

So the series converges absolutely for $\operatorname{Re} s > 3$.

• I don't know if such exercise are in the literature. – user243301 Feb 18 '16 at 21:06
• If you can find an $\alpha$ so that $S(n) \in \Theta(n^{\alpha})$, determining the abscissa of absolute convergence is easy. – Daniel Fischer Feb 18 '16 at 21:11
• Well, I don't know how get $O(n^\alpha)$, but you can tell me how if we assume that $S(n)\in O(n^\alpha)$, how find the abscissa of absolute convergence, so I know for the next time. Thanks @DanielFischer – user243301 Feb 18 '16 at 21:14
• If $\lvert a_n\rvert \leqslant C\cdot n^{\alpha}$, then absolute convergence of $\sum \frac{n^{\alpha}}{n^s}$ - which happens if and only if $\operatorname{Re} s > \alpha + 1$ implies the absolute convergence of $$\sum_{n = 1}^{\infty} \frac{a_n}{n^s}.$$ Conversely, if $c\cdot n^{\beta} \leqslant a_n$, the absolute convergence of $\sum \frac{a_n}{n^s}$ implies the absolute convergence of $\sum \frac{n^{\beta}}{n^s}$. So if $a_n \in \Theta(n^{\alpha})$, the abscissa of absolute convergence is $\alpha + 1$. – Daniel Fischer Feb 18 '16 at 21:22
• Very thanks much for the deduction and write the proposition @DanielFischer – user243301 Feb 18 '16 at 21:24

Answer for question 1: Following (I copy these) the comments written below myself question, by the user, for our coefficients $S(n)$, we have clearly $$S(n)\leq \sum_{k=1}^n k,$$ so we have $S(n)\in O(n^2)$. We can also find a lower bound for $S(n)$ of the form $c\cdot n^{\alpha}$, when we answer the question what is the largest $\alpha$ that we can reach? Well, if we're only looking at $k \geq n/2$, then $n \bmod k = (n-k)$, and summing (for example with Gauss identity) that gives approximately $n^2/8$. By calculus, see comments about the computations of the abscissa of absolute convergence, since we show that $S(n) \in \Theta(n^{\alpha})$, then we can say that the abscissa of absolute convergence is $\alpha+1=3$ .So the series converges absolutely for $\operatorname{Re} s > 3$.

My answer for question 2: Then by Lemma 1 (page 226 of Apostol's book) for $\sigma\geq c>3$ and $N>\geq 1$ $$|\sum_{n=N}^\infty\frac{S(n)}{n^s}|\leq N^{-(\sigma-c)}C\cdot\sum_{n=N}^{\infty}\frac{1}{n^{c-2}}.$$

I know that I can write (from the asymptotic formulas derived using Euler's summation, stated as Theorem 3.2 in page 55)

$$\sum_{n=N}^\infty\frac{1}{n^{c-2}}=O(N^{3-c}).$$

I don't say nothing more best than this. I don't try compute approximations for these tails to do a comparision.