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 $\operatorname{Re} s > 3$.
 A: 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. 
If you can improve my answer your are welcome.
