TITLE: Is $\sum_{n=1}^\infty\frac{S(n)}{n!}$ an irrational, where $S(n)$ denotes the sum of remainders function?
QUESTION [12 upvotes]: For each integer $n\geq 1$ we consider the arithmetic function $$S(n)=\sum_{k=1}^n n\text{ mod }k,\tag{1}$$
the sum of remainders function, the arithmetic function A004125 from the OEIS.
Example. We've that for $n=6$ $$S(6)=0+0+0+6\text{ mod }4+6\text{ mod }5+0=2+1=3.$$
This arithmetic function was studied for example in [1]. I wondered about a type of problems that are in the literature, that is in our case what about the irrationality of the real number $$\sum_{n=1}^\infty\frac{S(n)}{n!}.\tag{2}$$
I don't know if this previous example is in the literature or has good mathematical content.

Question. Is it possible to deduce that $$\sum_{n=1}^\infty\frac{S(n)}{n!}$$
  is irrational? Or well, is it possible to discard it as an irrational? 

I am asking about if it possible to do or provide some work, reasonings or heuristics, about it. Then I should to accept an answer. If it is in the literature refer the article and I try to search and read the statement and proof.
References:
[1] Michael Z. Spivey, The Humble Sum of Remainders Function, Mathematics Magazine, Vol. 78, No. 4 (Oct., 2005).

REPLY [4 votes]: Well, let me elaborate Ilya Bogdanov's argument. First of all, $$S(n)=\sum_{k=1}^n \left(n-k\lfloor n/k\rfloor\right)=n^2-\sum_{k,d:kd\leqslant n} k=
n^2-\sum_{d=1}^n (1+2+\ldots+\lfloor n/d\rfloor).
$$ 
We have $1+2+\ldots+\lfloor n/d\rfloor=\frac1{2d^2}n^2+O(n/d)$, thus
$$S(n)=\beta n^2+O\left(n\sum_{i=1}^n \frac1d+n^2\sum_{d=n+1}^\infty\frac1{2d^2}\right)=\beta n^2+O(n\log n),$$
where $\beta=1-\frac12\sum_{d=1}^\infty \frac1{d^2}=\frac12-\frac{\pi^2}{12}$. Assume that $\alpha:=\sum_{k=1}^\infty \frac{S(k)}{k!}$ is rational.
Choose large $n$. Then
$$(n-1)!\alpha=(n-1)!\sum_{k=1}^\infty \frac{S(k)}{k!}=\text{integer}+\frac{S(n)}n+\frac{S(n+1)}{n(n+1)}+o(1)=\\ \text{integer}+\frac{S(n)}n+\beta+o(1),$$
thus $S(n)/n=k_n-\beta+o(1)$ for certain integer $k_n$. We get
$$
k_{n+1}-\beta+o(1)=\frac{S(n+1)}{n+1}=\frac{S(n)+(2n+1)-\sigma(n+1)}{n+1}=\\
\frac{S(n)}{n+1}+\frac{(2n+1)-\sigma(n+1)}{n+1}=
\frac{S(n)}n-\frac{S(n)}{n(n+1)}+\frac{(2n+1)-\sigma(n+1)}{n+1}=\\
k_n-\beta+o(1)-\beta+\frac{(2n+1)-\sigma(n+1)}{n+1}.
$$
If, say, $n+1$ is large prime, this is not possible.