Another limit involving the fractional part

Question

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.

Answer 1

The conjecture is true. For $t\in[0,1]$, let $N(t)$ be the number of $k\in\{1,2,\dotsc,n\}$ satisfying $\{n/k\}<t$. On the one hand, $$\sum_{k=1}^n f\left(\frac{n}{k}\right)=\int_0^{1/2}\bigl(N(t+1/2)-N(t)\bigr)\,dt.$$ On the other hand, for any positive integer $M$, $$N(t)=\sum_{m=1}^\infty\left(\biggl\lfloor\frac{n}{m}\biggr\rfloor-\biggl\lfloor\frac{n}{m+t}\biggr\rfloor\right)=\sum_{m=1}^M\left(\frac{n}{m}-\frac{n}{m+t}\right)+O\left(M+\frac{n}{M}\right).$$ It follows that \begin{align*} \sum_{k=1}^n f\left(\frac{n}{k}\right) &=\sum_{m=1}^M\int_0^{1/2}\left(\frac{n}{m+t}-\frac{n}{m+t+1/2}\right)\,dt+O\left(M+\frac{n}{M}\right)\\ &=\sum_{m=1}^M n\log\left(\frac{(2m+1)^2}{(2m)(2m+2)}\right)+O\left(M+\frac{n}{M}\right)\\ &=n\log\left(\prod_{m=1}^M\frac{(2m+1)^2}{(2m)(2m+2)}\right)+O\left(M+\frac{n}{M}\right). \end{align*} By Wallis's product, $$\prod_{m=1}^\infty\frac{(2m+1)^2}{(2m)(2m+2)}=\frac{4}{\pi},$$ hence also $$\prod_{m=1}^M\frac{(2m+1)^2}{(2m)(2m+2)}=\frac{4}{\pi}\prod_{m=M+1}^\infty\frac{(2m)(2m+2)}{(2m+1)^2}=\frac{4}{\pi}\left(1+O\left(\frac{1}{M}\right)\right).$$ Taking the logarithm of both sides, and going back to the $k$-sum, $$\sum_{k=1}^n f\left(\frac{n}{k}\right)=n\log\left(\frac{4}{\pi}\right)+O\left(M+\frac{n}{M}\right).$$ Finally, we choose $M=\lfloor\sqrt{n}\rfloor$ to conclude that $$\sum_{k=1}^n f\left(\frac{n}{k}\right)=n\log\left(\frac{4}{\pi}\right)+O\left(\sqrt{n}\right).$$

Answer 2

Here are more details on my comment. Let's admit that we have $$(1)\,\,\,\, \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{n}{k}\right)=\int_{0}^ {1}f\left(\frac{1}{t}\right)dt$$

Since $f(x)=\frac{1}{2}\left(1-\left|2\left\{ x\right\} -1\right|\right)$ we have

$$\int_{0}^{1}f\left(\frac{1}{t}\right)dt=\int_{1}^{\infty}f(u)\frac{du}{u^{2 }}=\frac{1}{2}\sum_{k=1}^{\infty}\int_{k}^{k+1}\left(1-\left|2(u-k)-1\right |\right)\frac{du}{u^{2}}$$

$$=\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}+\int_{k}^ {k+1/2}\left(2u-2k-1\right)\frac{du}{u^{2}}-\int_{k+1/2}^{k+1}\left(2u -2k-1\right)\frac{du}{u^{2}}\right)$$

$$=\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}-\frac{1}{ k}-2\log\left(\frac{k}{k+1/2}\right)+\frac{1}{k+1}-2\log\left(\frac{k+1}{ k+1/2}\right)\right)$$

$$=\log\left(\prod_{k=1}^{\infty}\frac{(k+1/2)^{2}}{k(k+1)}\right)=\log\left( \frac{4}{\pi}\right)$$