It is known that

$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\left\{ \frac{n}{k}\right\} =1-\gamma$$

where $\left\{ x\right\}$ is the fractional part of $x$ and $\gamma$ is the Euler constant. Let $f(x)=\min\left(\left\{ x\right\} ,1-\left\{ x\right\} \right)$. Do we have

$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{n}{k}\right)=\log\left(\frac{4}{\pi}\right)=0.2415644752...?$$

the “alternating Euler constant” (cf. https://oeis.org/A094640)? For instance I get for $n=10^{8}$

$\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{n}{k}\right)=0.2415641681...$

And if this is the case can we hope that

$$\sum_{k=1}^{n}f\left(\frac{n}{k}\right)=\log\left(\frac{4}{\pi}\right)n+O\left( n^{1/2}\right)?$$ which seems reasonable when carrying out numerical tests.