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> For two sequences $a\_n$ and $b\_n$, “If $\{a\_n\}$ and $\{b\_n\}$ are increasing, then $\{a\_nb\_n\}$ is increasing.” Show this is false, make the hypothesis on $\{b\_n\}$ stronger, and prove the amended statement.
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I was thinking to let both $a\_n$ and $b\_n$ be positive, but it only let me change the hypothesis on $b\_n$, how do I proceed?
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Suppose that $\{a\_n\}$ is increasing, $\{b\_n\}$ is positive and increasing, and that $\{b\_n\}$ also has the following property:

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> for all $n$, if $a\_{n+1}<0$ then $b\_{n+1}\le a\_nb\_n/a\_{n+1}$.
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First check that the conditions on $\{b\_n\}$ are not inconsistent: if $a\_{n+1}<0$ then $a\_n\le a\_{n+1}<0$, so $a\_nb\_n/a\_{n+1}\ge b\_n$ and hence it is possible to choose $b\_{n+1}$ satisfying $b\_n\le b\_{n+1}\le a\_nb\_n/a\_{n+1}$.

Now if $a\_{n+1}<0$ then this condition clearly gives
$$a\_{n+1}b\_{n+1}\ge a\_nb\_n\ ;$$
if on the other hand $a\_{n+1}\ge0$ then
$$a\_{n+1}b\_{n+1}-a\_nb\_n=a\_{n+1}(b\_{n+1}-b\_n)+b\_n(a\_{n+1}-a\_n)\ge0\ .$$
It's possible that there may be an error in the exercise; if you require that the new condition on $b\_n$ be independent of $a\_n$, then you have a problem: For all increasing sequences $b\_n$, there exists a sequence $a\_n$ such that $a\_nb\_n$ is not increasing. One can prove this by cases: If $b\_1$ and $b\_2$ are positive, then set $a\_1=-1$ and $a\_2=-\frac{1}2\left(\frac{b\_1}{b\_2}+1\right)$. If both $b\_1$ and $b\_2$ are negative, set $a\_1=1$ and $a\_2=\frac{b\_1}{b\_2}+1.$ If $b\_1$ is negative and $b\_2$ is positive, set $a\_1=-1$ and $a\_2=0$.

If we weaken the condition on the sequences to be "non-decreasing" then at least we could use "$b\_n$ is constant and non-negative". If we allow $b\_n$ to depend on $a\_n$, then we get trivial conditions like "$b\_n$ is such that $a\_nb\_n$ is increasing". None of these conditions seem terribly natural though.