It is known that
lim
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.