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!
 A: 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 $n=10^{100}$.
