A series: $1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\cdots$

Denote $$b_1=1,b_{n}=b_{n-1}-\dfrac{S(b_{n-1})}{n},(n>1 )\tag1$$ where $S(x)=1$ if $x>0,S(x)=-1$ if $x<0$, and $S(0)=0.$

So $b_n=1,\dfrac{1}{2},\dfrac{1}{6},-\dfrac{1}{12},\dfrac{7}{60},-\dfrac{1}{20},\dfrac{13}{140},-\dfrac{9}{280},\dfrac{199}{2520},-\dfrac{53}{2520}…$

For example: $b_7=1-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{7}=\dfrac{13}{140}.$

We can prove that $$\lim_{n \rightarrow \infty} b_n=0,$$ and $$\sum_{i=0}^2{S(b_{n+i})}=±1. (n>1) \tag2$$

Question 1: Can we determine $S(b_n)$ for very large $n$ (such as $n=10^{100}$)?

Question 2: Prove or disprove that if $n>3$, then $$\left|\sum_{i=4}^n{S(b_i)}\right|\leq 1.\tag3$$

I have checked it for $n<4 \cdot 10^4.$Has this problem been studied before?Thanks in advance!

I can't answer question 1, but I do have an answer for question 2. In particular, the fact that $$ \left|\sum_{i=4}^n S(b_i)\right| \leq 1 $$ is an immediate consequence of the following theorem.

Theorem. For all $n\geq 2$, the terms $b_{2n}$ and $b_{2n+1}$ have opposite signs.

Proof: It is easy to check that the theorem holds for $2\leq n\leq 7$. Therefore, it suffices to show that $$ |b_{2n}| < \dfrac{1}{2n+1} $$ for all $n\geq 8$. To do so, we shall prove the much better bound $$ |b_{2n}| \;\leq\; \frac{1}{2n-1}-\frac{1}{2n} $$ for all $n\geq 8$. (Note that this difference is always less than $1/(2n+1)$.)

We proceed by induction on $n$. The base cases is $n=8$, for which $$ |b_{16}| \;=\; \frac{547}{144144} \;\leq\; \frac{1}{15}-\frac{1}{16}. $$ Now suppose that $|b_{2n}|\leq \dfrac{1}{2n-1}-\dfrac{1}{2n}$. Without loss of generality, we may assume that $b_{2n}$ is positive. Then $b_{2n} \leq \dfrac{1}{2n+1}$, so $$ b_{2n+1} \;=\; b_{2n} \,-\, \frac{1}{2n+1} $$ is negative, and therefore $$ b_{2n+2} \,=\, b_{2n+1} + \frac{1}{2n+2} \,=\, b_{2n} - \frac{1}{2n+1} + \frac{1}{2n+2}. $$ But $b_{2n} \leq \dfrac{1}{2n-1}-\dfrac{1}{2n} \leq 2\left(\dfrac{1}{2n+1} - \dfrac{1}{2n+2}\right)$, so it follows that $$ |b_{2n+2}| \;\leq\; \frac{1}{2n+1} - \frac{1}{2n+2}.\tag*{$\square$} $$

This proof was based on the observation that $b_n$ is "small" when $n$ is even, and "large" when $n$ is odd. There are similar patterns for other powers of $2$: the term $b_n$ tends to be "small" when $n$ is $2$ mod $4$, and when $n$ is $6$ mod $8$.

Indeed, the theorem above appears to be just the first in a sequence of theorems about the signs of the $b_n$. Then next two are

Theorem. For all $n\geq 4$, the terms $b_{4n+2}$ and $b_{4n+4}$ have opposite signs.

Theorem. For all $n\geq 2$, the terms $b_{8n+6}$ and $b_{8n+10}$ have opposite signs.

The first of these follows from the fact that $$ |b_{4n+2}| \;\leq\; \frac1{4 n - 1} - \frac1{4 n} - \frac1{4 n + 1} + \frac1{4 n + 2} $$ for $n\geq 5$. The second follows from the fact that $$ |b_{8n+6}| \;\leq\; \frac1{8 n - 1} - \frac1{8 n} - \frac1{8 n + 1} + \frac1{8 n + 2} - \frac{1}{8n+3} + \frac{1}{8n+4} +\frac{1}{8n+5} - \frac{1}{8n+6} $$ for all $n\geq 2$.

What seems to be happening is that it gets "caught" in these patterns, e.g. if the $k$th term happens to be unusually small (on the order of $1/n^{2^n}$), then the same will be true of the $(k+2^n)$'th term. However, I don't see any way to predict when it gets "caught", or what the parity will be (i.e. why 6 mod 8 instead of 2 mod 8), so I don't have any way of predicting the sign of $b_n$ for $n=10^{100}$.