<s>[INST] If $a_n\sim b_n$ decrease to $0$ and $a_{n+1}b_n-a_nb_{n+1}$ changes sign i.o. , does it imply the same for $(1-a_{n+1})(1-b_n)-(1-a_n)(1-b_{n+1})$? 
Suppose $a_n<b_n$ for all $n$, both are strictly decreasing to $0$ and $a_n\sim b_n$. If $A_n=a_{n+1}b_n-a_nb_{n+1}$ changes sign infinitely often, does it follow that $B_n=(1-a_{n+1})(1-b_n)-(1-a_n)(1-b_{n+1})$ also does?

In some sense, the implication should not be there, because while $A_n<0$ does not seem to be crucial for $B_n$, I have found that whenever $A_n>0$ we have $B_n<0$: the former is equivalent to $$a_{n+1}>a_n\frac{b_{n+1}}{b_n}$$ and since for large enough $n$, $0<a_n<1, a_n<b_n<1, 0<b_{n+1}<b_n$ this is stronger than $$a_{n+1}>1-\frac{1-a_n}{1-b_n}(1-b_{n+1}),$$ equivalent to $B_n<0$. $A_n<0$ does not seem to be crucial because it should be possible to have this happen without making $a_{n+1}$ so small that $B_n$ becomes positive.
On the other hand, maybe the fact that $A_n$ changes sign infinitely often gives more weight to the instances of $A_n<0$. Not for too long, but I have tried to construct some simple $A_n,B_n$ contradicting the claim and I have not found them yet. Are there any, with $B_n<0$? Besides, how does the situation change if $a_n-b_n$ is also assumed to change sign infinitely often?
 [/INST] $\def\peq{\mathrel{\phantom{=}}{}}$Take an arbitrary $c > 1$ and define$$
a_n = \frac{1}{c^n + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2}}}, \quad b_n = \frac{1}{c^n}. \quad \forall n \geqslant 1
$$
It is easy to verify that $a_n < b_n$, $a_n \sim b_n\ (n → ∞)$, and both $\{a_n\}$ and $\{b_n\}$ strictly decrease to $0$. Now define$$
x_n = \frac{1}{a_n} = c^n + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2}}, \quad y_n = \frac{1}{b_n} = c^n, \quad z_n = x_n - y_n,\quad \forall n \geqslant 1
$$
then\begin{align*}
A_n > 0 &\Longleftrightarrow \frac{x_{n + 1}}{x_n} < \frac{y_{n + 1}}{y_n} \Longleftrightarrow \frac{c^{n + 1} + c^{\frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2}}}{c^n + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2}}} < c\\
&\Longleftrightarrow c^{n + 1} + c^{\frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2}} < c^{n + 1} + c^{\frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2} + 1}\\
&\Longleftrightarrow \frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2} < \frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2} + 1\\
&\Longleftrightarrow n \text{ is even}.
\end{align*}
and\begin{align*}
B_n < 0 &\Longleftrightarrow (a_n - a_{n + 1}) - (b_n - b_{n + 1}) + (a_{n + 1} b_n - a_n b_{n + 1}) < 0\\
&\Longleftrightarrow y_n y_{n + 1}(x_{n + 1} - x_n) - x_n x_{n + 1}(y_{n + 1} - y_n) + (x_n y_{n + 1} - x_{n + 1} y_n) < 0.
\end{align*}
Since\begin{align*}
&\peq y_n y_{n + 1}(x_{n + 1} - x_n) - x_n x_{n + 1}(y_{n + 1} - y_n) + (x_n y_{n + 1} - x_{n + 1} y_n)\\
&= y_n y_{n + 1}\bigl( (y_{n + 1} - y_n) - (z_{n + 1} - z_n) \bigr) - (y_n + z_n)(y_{n + 1} + z_{n + 1})(y_{n + 1} - y_n)\\
&\peq + \bigl( (y_n + z_n) y_{n + 1} - (y_{n + 1} + z_{n + 1}) y_n \bigr)\\
&= \color{blue}{y_n y_{n + 1}}\bigl( \color{red}{(y_{n + 1} - y_n)} - (z_{n + 1} - z_n) \bigr) - (\color{blue}{y_n y_{n + 1}} + y_n z_{n + 1} + y_{n + 1} z_n + z_n z_{n + 1})\color{red}{(y_{n + 1} - y_n)}\\
&\peq + \bigl( (y_n + z_n) y_{n + 1} - (y_{n + 1} + z_{n + 1}) y_n \bigr)\\
&= y_n y_{n + 1} (z_{n + 1} - z_n) - (y_n z_{n + 1} + y_{n + 1} z_n + z_n z_{n + 1})(y_{n + 1} - y_n) + (y_{n + 1} z_n - y_n z_{n + 1})\\
&= y_n y_{n + 1} (z_{n + 1} - z_n) - (y_n z_{n + 1} + y_{n + 1} z_n) y_{n + 1} + (y_n z_{n + 1} + y_{n + 1} z_n) y_n\\
&\peq - z_n z_{n + 1}(y_{n + 1} - y_n) + (y_{n + 1} z_n - y_n z_{n + 1})\\
&= y_n^2 z_{n + 1} - y_{n + 1}^2 z_n - z_n z_{n + 1}(y_{n + 1} - y_n) + (y_{n + 1} z_n - y_n z_{n + 1})\\
&= (y_n^2 - y_n) z_{n + 1} - (y_{n + 1}^2 - y_{n + 1}) z_n - z_n z_{n + 1}(y_{n + 1} - y_n),
\end{align*}
and\begin{gather*}
\frac{y_{n + 1}^2 - y_{n + 1}}{y_n^2 - y_n} = \frac{y_{n + 1}}{y_n} \cdot \frac{y_{n + 1} - 1}{y_n - 1} > c^2,\\
\frac{z_{n + 1}}{z_n} = c^{\left( \frac{3}{4} \left[ \frac{n + 1}{2} \right] + \frac{n + 1}{2} \right) - \left( \frac{3}{4} \left[ \frac{n}{2} \right] + \frac{n}{2} \right)} = c^{\frac{3}{4} \left( \left[ \frac{n + 1}{2} \right] - \left[ \frac{n}{2} \right] \right) + \frac{1}{2}} < c^2\\
\Longrightarrow \frac{y_{n + 1}^2 - y_{n + 1}}{y_n^2 - y_n} > \frac{z_{n + 1}}{z_n} \Longrightarrow (y_n^2 - y_n) z_{n + 1} - (y_{n + 1}^2 - y_{n + 1}) z_n < 0,
\end{gather*}
then $B_n < 0$ for all $n \geqslant 1$.
</s>