$\newcommand{\pars}[1]{\left\{ \frac{n}{#1} \right\}}$

$$\lim_{n\to\infty}\dfrac{1}{n} \left( \pars{1} - \pars{2} + ... + (-1)^{n+1} \pars{n} \right),$$

where $\left\{ x \right\} $ denotes the fractional part of $x$.

My guess is that the limit is equal to 0; I tried finding some asymptotics for the fractional part sum, by looking for example at ways to bound $\pars{k} - \pars{k+1}$; My intuition is that this difference is rather small(perhaps less than $\frac{n}{k(k+1)}$) and that it is big enough to be relevant only when one of them is zero, meaning that $k$ or $k+1$ divides $n$. This would lead me to conjecture that it grows at most as $O(\sqrt{n})$, which would make the limit zero, but I have not been able to make this rigorous. Another idea I had would be to look at the sum with odd denominator and the sum with even denominators and show that they must be "rather" close; this seems pretty intuitive but the fractional part is very chaotic and I have not been able to get any bounds.

Any ideas/tips would be appreciated!


Yes, the limit is zero. Let's denote $n_k=\lfloor n/k\rfloor$. Then, for $1\leqslant r\leqslant n$, we have $$n_k=r\iff kr\leqslant n<k(r+1)\iff n_{r+1}<k\leqslant n_r.\tag{*}\label{basics}$$ In particular, for "small" $r$, there are "long" runs of $k$ with the same value of $n_k(=r)$.

This leads to a solution. Let $1<m<n$. Splitting the sum $\sum\limits_{k=1}^{n}=\sum\limits_{k=1}^{n_m}+\sum\limits_{r=2}^{m}\sum\limits_{k=n_r+1}^{n_{r-1}}$, we get, crudely, $$\left|\sum_{k=1}^{n}(-1)^{k-1}\left\{\frac{n}{k}\right\}\right|\leqslant\underbrace{\left|\sum_{k=1}^{n_m}(-1)^{k-1}\left\{\frac{n}{k}\right\}\right|}_{<n_m}+\sum_{r=2}^{m}\underbrace{\left|\sum_{k=n_r+1}^{n_{r-1}}(-1)^{k-1}\left\{\frac{n}{k}\right\}\right|}_{<m+n/n_m};$$ the second estimate holds because, under that sum, we have $\lfloor n/k\rfloor=r-1$ (see \eqref{basics}), hence $$\Bigg|\sum_{k=n_r+1}^{n_{r-1}}(-1)^{k-1}\underbrace{\left(\frac{n}{k}-r+1\right)}_{=\{n/k\}}\Bigg|\leqslant n\Bigg|\sum_{k=n_r+1}^{n_{r-1}}\frac{(-1)^{k-1}}{k}\Bigg|+(r-1)\Bigg|\sum_{k=n_r+1}^{n_{r-1}}(-1)^{k-1}\Bigg|<\frac{n}{n_r}+r.$$

Thus, we get an estimate $$\left|\sum_{k=1}^{n}(-1)^{k-1}\left\{\frac{n}{k}\right\}\right|<n_m+m^2+\frac{mn}{n_m},$$ and, taking $m=\lfloor n^{1/3}\rfloor$, we see that this is $\mathcal{O}(n^{2/3})$, which is sufficient.

This leaves the question whether there is a better estimate [than $\mathcal{O}(n^{2/3})$] open. Any refinements?..

  • 1
    $\begingroup$ The asymptotic of $D(n)=\sum_{k\le n} \lfloor n/k\rfloor$ is the Dirichlet divisor problem, here it is $\sum_{k\le n} (-1)^k n/k-(D(n)-2D(n/2) )$ which is the same. I'd say doing better than $O(n^{1/2})$ needs the analytic methods in Titchmarsh $\endgroup$ – reuns Jan 17 '20 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.