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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.

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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    $\begingroup$ @YoTengoUnLCD No it doesn't; if $a$ starts with $-2,-1$ and $b$ starts with $1,4$, then $ab$ starts as $-2,-4$, which isn't increasing. $\endgroup$ Feb 7 '16 at 23:52
  • $\begingroup$ Woops, true. My bad. $\endgroup$ Feb 7 '16 at 23:55
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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:

for all $n$, if $a_{n+1}<0$ then $b_{n+1}\le a_nb_n/a_{n+1}$.

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 $$a_{n+1}b_{n+1}-a_nb_n=a_{n+1}(b_{n+1}-b_n)+b_n(a_{n+1}-a_n)\ge0\ .$$

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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.

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