I have been noodling around with the function
$$f(x):=\frac{\sum _{j=1}^x \text{frac}\left(\frac{x}{j}\right)}{x}$$
where $x$ is a positive integer, and $\text{frac}(n)$ denotes the fractional part of $n$. I initially thought that $f$ would be asymptotic to $x=\frac{x}{2}$ since for large $x$ each of the possible values
$$\text{frac}\left(\frac{x}{j}\right)=\left\{0,\frac{1}{j},\frac{2}{j},\ldots ,\frac{j-1}{j}\right\}$$
should occur with equal probability. However, my assumption is clearly wrong. Heuristically, the asymptote is in fact something like $x=\frac{21}{50}$, as the plot below shows.
Can someone please explain why?
By the known asymptotics of the harmonic numbers and the divisor summatory function, we have \begin{align*} \sum\limits_{j = 1}^x {\left\{ {\frac{x}{j}} \right\}} & = \sum\limits_{j = 1}^x {\frac{x}{j}} - \sum\limits_{j = 1}^x {\left\lfloor {\frac{x}{j}} \right\rfloor } = x\sum\limits_{j = 1}^x {\frac{1}{j}} - \sum\limits_{j = 1}^x {d(j)} \\ & = x\left( {\log x + \gamma + \mathcal{O}\!\left( {\frac{1}{x}} \right)} \right) - x\log x - x(2\gamma - 1) + \mathcal{O}(\sqrt x ) \\ & = x(1-\gamma)+\mathcal{O}(\sqrt x ). \end{align*} Thus $$ \frac{1}{x}\sum\limits_{j = 1}^x {\left\{ {\frac{x}{j}} \right\}} = 1 - \gamma + \mathcal{O}\!\left( {\frac{1}{{\sqrt x }}} \right). $$ This shows that the limit is $1-\gamma =0.4227843350\ldots$.