$\lim_{n\to\infty}\frac{n -\big\lfloor\frac{n}{2}\big\rfloor+\big\lfloor\frac{n}{3}\big\rfloor-\dots}{n}$, a Brilliant problem

Let $f : [0, 1] \to \mathbb{R}$ be defined by

$$f(x) = \mathbf{1}_{\{\text{$x > 0$ and $\lfloor 1/x \rfloor$ is odd}\}} = \sum_{i=1}^{\infty} \mathbf{1}_{\{ 2i-1 \leq \frac{1}{x} < 2i \}}. $$

Then by double counting, we find that

\begin{align*} s_n = \sum_{k=1}^{n} (-1)^{k-1} \bigg\lfloor \frac{n}{k} \bigg\rfloor &= \sum_{k=1}^{n} (-1)^{k-1} \sum_{j=1}^{n} \mathbf{1}_{\{kj \leq n\}} \\ &= \sum_{j=1}^{n} \sum_{k=1}^{n} (-1)^{k-1} \mathbf{1}_{\{kj \leq n\}} = \sum_{j=1}^{n} f\left(\frac{j}{n}\right). \end{align*}

Now we utilize the following lemma:

Lemma. Let $f : [0, 1] \to \mathbb{R}$ be Riemann integrable. Then $$ \lim_{n\to\infty} \sum_{j=1}^{n} f\left(\frac{j}{n}\right)\frac{1}{n} = \int_{0}^{1} f(x) \, dx. $$

From this, we know that $s_n/n$ converges and

$$ \lim_{n\to\infty} \frac{s_n}{n} = \int_{0}^{1} f(x) \, dx = \sum_{i=1}^{\infty} \left( \frac{1}{2i-1} - \frac{1}{2i} \right) = \log 2. $$

Note that $$\frac{1}{n}\sum_{k=1}^{n}(-1)^{k+1}\left\lfloor\frac{n}{k}\right\rfloor=\frac{1}{n}\sum_{k=1}^{n}\left\lfloor\frac{n}{k}\right\rfloor-\frac{1}{n/2}\sum_{k=1}^{\lfloor n/2\rfloor}\left\lfloor\frac{n/2}{k}\right\rfloor=\frac{D(n)}{n}-\frac{D(n/2)}{n/2}.$$ where $D(x)= \sum_{k\geq 1}^{n}\left\lfloor\frac{x}{k}\right\rfloor$ is the divisor summatory function. It is known that $$D(x) = x\ln(x) + x(2\gamma -1) + O(\sqrt{x})\implies \frac{D(x)}{x} = \ln(x) + (2\gamma -1) + o(1)$$ (see also Dirichlet's Divisor Problem). Hence, as $n$ goes to $+\infty$, $$\frac{1}{n}\sum_{k=1}^{n}(-1)^{k+1}\left\lfloor\frac{n}{k}\right\rfloor= \ln(n)-\ln(n/2)+o(1)\to \ln(2).$$