Is $\sum_{n=1}^\infty\frac{S(n)}{n!}$ an irrational, where $S(n)$ denotes the sum of remainders function?

Question

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).

Answer 1

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.

Answer 2

This is just a reduction of the claim to an other (plausible) claim.

Notice that $S(n)\leq 1+2+\dots+\lfloor n/2\rfloor+\lceil n/2\rceil+\dots+1=n^2/4+O(n)$. This yields $$ \sum_{k>n}\frac{S(k)}{k!}\leq \frac1{(n-1)!}\left(\frac12+o(1)\right). $$

Now, assume that $$ S(n)\mod n<\frac{n}{2+\varepsilon} \qquad\text{for infinitely many $n$.}\qquad(*) $$ For such $n$ being large enough, denoting the sum under consideration by $\alpha$, we would have $$ \alpha=\frac{A}{(n-1)!}+\frac{S(n)\mod n}{n!}+\sum_{k>n}\frac{S(k)}{k!} \in\left(\frac{A}{(n-1)!},\frac{A+1}{(n-1)!}\right) $$ for some integer $A$. Therefore, $\alpha$ is not a ratio of an integer with $(n-1)!$ (for infinitely many $n$), which means $\alpha$ is irrational.

Thus, in order to show the irrationality of $\alpha$, it sufices to prove $(*)$.

In particular, it would suffice to prove that there are infinitely many $n$ with $n\mid S(n)$...

Remark. In fact, $$ \frac{S(n)}{n^2}\to \sum_{k=1}^\infty \frac1{2k(k+1)^2} \qquad\text{as $n\to\infty$.} $$ Does it help?

Answer 3

Here's a partial answer (too long for a coment) that attempts to evaluate the sum $$\sum_{n=1}^\infty\frac{S(n)}{n!}$$ directly. Let $\sigma(n)$ denote the sum of divisors of $n$. First note that $$S(n)=S(n-1)+(n-1)-(\sigma(n)-n)$$ because each of the $n-1$ remainders increases by $1$ but there is overflow whenever $k$ is a proper divisor of $n$ (and the sum of such $k$'s is $\sigma(n)-n$). Since $S(0)=0$, this gives the formula $$S(n)=\sum_{k=1}^n((2k-1)-\sigma(k))=n^2-\sum_{k=1}^n\sigma(k)=n^2-\sum_{d=1}^nd\left\lfloor\frac{n}{d}\right\rfloor.$$ Then $$\sum_{n=1}^\infty\frac{S(n)}{n!}=\sum_{n=1}^\infty\frac{n^2}{n!}-\sum_{n=1}^\infty\sum_{d=1}^n\frac{d}{n!}\left\lfloor\frac{n}{d}\right\rfloor=2e-\sum_{d=1}^\infty\sum_{n=0}^\infty\frac{d}{n!}\left\lfloor\frac{n}{d}\right\rfloor.$$ If we let $n\%d$ be the remainder of $n$ when dividing by $d$, then $$\sum_{n=0}^\infty\frac{d}{n!}\left\lfloor\frac{n}{d}\right\rfloor=\sum_{n=0}^\infty\frac{n-(n\%d)}{n!}=e-\sum_{n=0}^\infty\frac{n\%d}{n!}.$$ In this last sum, the coefficients of $\frac{1}{n!}$ have period $n$. Then abstractly, this last sum should be a formal linear combination of the $n$ power series for $e^\omega$ as $\omega$ runs through the $n$th roots of unity. Indeed, $$\sum_{n=0}^\infty\frac{n\%d}{n!}=\frac{d-1}{2}e+\sum_{k=1}^{d-1}\frac{1}{\omega_d^{-k}-1}e^{\omega_d^k}$$ where $\omega_d$ is a primitive $d$th root of unity so $$\sum_{n=0}^\infty\frac{d}{n!}\left\lfloor\frac{n}{d}\right\rfloor=\frac{3-d}{2}e-\sum_{k=1}^{d-1}\frac{1}{\omega_d^{-k}-1}e^{\omega_d^k}.$$ Thus, \begin{align*} \sum_{n=1}^\infty\frac{S(n)}{n!}&=2e-\sum_{d=1}^\infty\left(\frac{3-d}{2}e-\sum_{k=1}^{d-1}\frac{1}{\omega_d^{-k}-1}e^{\omega_d^k}\right)\\ &=e-\sum_{d=2}^\infty\left(\frac{3-d}{2}e-\sum_{k=1}^{d-1}\frac{1}{\omega_d^{-k}-1}e^{\omega_d^k}\right)\\ \end{align*} The number-theoretic functions are all gone but unfortunately, this still looks rather intractable. Maybe someone can see a way to sum this?