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Find

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

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

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!

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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?..

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    $\begingroup$ 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

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