Solution:

Define
$$f_k(x)=\sum_{n=0}^{2^k-1} \frac{(-1)^{s(n)}}{x+n}$$
We want to show that $f_{2022}(2022)>0$. We will in fact show something stronger.

I claim that for all $x>0$, for all $k \geq 0$, we have $f_k^{(i)}(x)>0$ for even $i$ and $f_k^{(i)}(x)<0$ for odd $i$, where $f^{(i)}$ denotes the $i$th derivative of $f$. We will prove this claim with induction on $k$.

The base case of $k=0$ is easy to see because $f_0(x)=\frac{1}{x}$, so $f_0^{(2j)}(x)=\frac{(2j)!}{x^{2j+1}}>0$ and $f_0^{(2j-1)}(x)=-\frac{(2j-1)!}{x^{2j}}<0$ for all $x>0$.

Now, assume the claim is true for $k=N$. Then, note that
$$\begin{gathered} f_{N+1}(x)=\sum_{n=0}^{2^{N+1}-1} \frac{(-1)^{s(n)}}{x+n}=\sum_{n=0}^{2^N-1} \frac{(-1)^{s(n)}}{x+n}+\sum_{n=0}^{2^N-1} \frac{(-1)^{s\left(n+2^N\right)}}{x+n+2^N}= \\ \sum_{n=0}^{2^N-1} \frac{(-1)^{s(n)}}{x+n}-\sum_{n=0}^{2^N-1} \frac{(-1)^{s(n)}}{x+n+2^N}=f_N(x)-f_N\left(x+2^N\right) \end{gathered}$$
Thus,
$$f_{N+1}^{(2j)}(x)=f_N^{(2j)}(x)-f_N^{(2j)}\left(x+2^N\right)>0$$
since $\left(f_N^{(2j)}(x)\right)'=f_N^{(2j+1)}(x)<0$. Similarly, we can show that $f_{N+1}^{2j+1}(x)<0$, which completes the induction, so we are done.

Solution:

Define the function
$$f(t)=t^{2021}(1-t)\left(1-t^{2}\right)\left(1-t^{4}\right)\left(1-t^{8}\right) \cdots\left(1-t^{2^{2021}}\right)=\sum_{n=0}^{2^{2022}-1}(-1)^{s(n)} t^{2021+n}$$
Note that we have $f(t)>0$ for all $t \in(0,1)$, so we have
$$0<\int_{0}^{1} f(t) d t=\sum_{n=0}^{2^{2022}-1}(-1)^{s(n)} \frac{1}{2022+n}$$
so we are done.