$\lim_{n\to\infty}\frac{n -\big\lfloor\frac{n}{2}\big\rfloor+\big\lfloor\frac{n}{3}\big\rfloor-\dots}{n}$, 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 $+\infty$, $$\frac{1}{n}\sum_{k=1}^{n}(-1)^{k+1}\left\lfloor\frac{n}{k}\right\rfloor= \ln(n)-\ln(n/2)+o(1)\to \ln(2).$$

Elementary high school approach

One can show immediately with the squeeze theorem that $$L_1=\lim_{n\to\infty}\left(\sum_{k=1}^{2\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)=\log(2),$$

using the simple fact that $\lim_{n\to\infty}(H_{2n}-H_n)=\log(2)$.

But guess what! Your limit is

$$\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{n}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)=L_1,$$ and therefore you can use the first limit to calculate your initial limit and then show that the sum of the remaining terms tends to $0$, which is straightforward.

Adding some steps requested by OP

WLOG, for the comfort of calculations, we can replace in the initial limit $n$ by $2n$, and using the double inequality with floor function $x\ge\left\lfloor x\right\rfloor\ge x-1$, we have

$$H_{2\left\lfloor\sqrt{n}\right\rfloor}\ge\sum_{k=1}^{2\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor\ge H_{2\left\lfloor\sqrt{n}\right\rfloor}-\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}$$

$$-H_{\left\lfloor\sqrt{n}\right\rfloor}+\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}\ge-\sum_{k=1}^{\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\ge-H_{\left\lfloor\sqrt{n}\right\rfloor},$$ that give $$H_{2\left\lfloor\sqrt{n}\right\rfloor}-H_{\left\lfloor\sqrt{n}\right\rfloor}+\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}\ge\sum_{k=1}^{2\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\ge H_{2\left\lfloor\sqrt{n}\right\rfloor}-H_{\left\lfloor\sqrt{n}\right\rfloor}-\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}.$$ Letting $n\to\infty$ you get the value of the limit $L_1$. If denoting $\left\lfloor\sqrt{n}\right\rfloor=m$, we have the type of limit mentioned above, $\lim_{m\to\infty}(H_{2m}-H_m)=\log(2)$.

Your limit, after replacing $n$ by $2n$, is

$$\lim_{n\to\infty}\sum_{k=1}^{2n}(-1)^{k-1}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-2\sum_{k=1}^{n}\frac{1}{2n}\left\lfloor\frac{2n}{2k}\right\rfloor\right)$$ $$=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{n}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right).$$

What's left to do? Using in the last limit that limit $L_1$ calculated above. $$\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{n}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)$$ $$=\lim_{n\to\infty}\left(\sum_{k=1}^{2\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor+\sum_{k=2\left\lfloor \sqrt{n} \right\rfloor+1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor-\sum_{k=\left\lfloor \sqrt{n} \right\rfloor+1}^n\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)$$ $$=\underbrace{\lim_{n\to\infty}\left(\sum_{k=1}^{2\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)}_{\displaystyle \log(2)}+\underbrace{\lim_{n\to\infty}\left(\sum_{k=2\left\lfloor \sqrt{n} \right\rfloor+1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=\left\lfloor \sqrt{n} \right\rfloor+1}^n\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)}_{\displaystyle 0}$$ $$=\log(2),$$ and the limit tending to $0$ can be done by arranging the sums under the limit under the form of an alternating sum and then squeezing the sum.

Further explanations on the limit tending to $0$

It's clear that $$0\le\sum_{k=2\left\lfloor \sqrt{n} \right\rfloor+1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=\left\lfloor \sqrt{n} \right\rfloor+1}^n\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor=\frac{1}{2n}\sum_{k=\underbrace{2\left\lfloor \sqrt{n} \right\rfloor+1}_{m}}^{2n}(-1)^{k-1}\left\lfloor\frac{2n}{k}\right\rfloor$$ $$=\frac{1}{2n}\left(\left\lfloor\frac{2n}{m}\right\rfloor-\left\lfloor\frac{2n}{m+1}\right\rfloor+\left\lfloor\frac{2n}{m+2}\right\rfloor-\left\lfloor\frac{2n}{m+3}\right\rfloor+\left\lfloor\frac{2n}{m+4}\right\rfloor-\left\lfloor\frac{2n}{m+5}\right\rfloor+\left\lfloor\frac{2n}{m+6}\right\rfloor-\cdots\right)$$ $$\le\frac{1}{2n}\left(\left\lfloor\frac{2n}{m}\right\rfloor-\left\lfloor\frac{2n}{m+1}\right\rfloor+\left\lfloor\frac{2n}{m+1}\right\rfloor-\left\lfloor\frac{2n}{m+3}\right\rfloor+\left\lfloor\frac{2n}{m+3}\right\rfloor-\left\lfloor\frac{2n}{m+5}\right\rfloor+\left\lfloor\frac{2n}{m+5}\right\rfloor-\cdots\right)$$ $$=\frac{1}{2n}\left(\left\lfloor\frac{2n}{m}\right\rfloor-1\right)=\frac{1}{2n}\left(\left\lfloor\frac{2n}{2\left\lfloor \sqrt{n} \right\rfloor+1}\right\rfloor-1\right),$$ where letting $n\to\infty$, we obviously get $0$ which accounts for the limit tending to $0$ in my solution.

A final remark

One should have dealt from the very beginning with this part of the limit tending to $0$ since we had it in the form with the alternating sum as expected.