Is the product of two monotone sequences monotone? Question: The product of monotone sequences is monotone, T or F?
Uncompleted Solution: There are four cases from considering each of two monotone sequences, increasing or decreasing.
CASE I: Suppose we have two monotonically decreasing sequences, say ${\{a_n}\}$ and ${\{b_n}\}$. Then, $a_{n+1}\leq a_n$ and $b_{n+1}\leq b_n$; if $b_n\geq 0$ and $b_{n+1}\geq 0$ then $a_{n+1}b_{n+1}\leq a_{n}b_{n+1}\leq a_{n}b_{n}$, but the l-h-s inequality, i.e., $a_{n}b_{n+1}\leq a_{n}b_{n}$, implies that must $a_{n}\geq 0$ since $b_{n+1}\leq b_n$ already has been supposed, but $a_{n}\geq 0$ has not been supposed. So does it mean that two monotonically decreasing sequences with requisites of $a_{n}\leq 0$ since $b_{n+1}\leq b_n$; is counterexample for "the product of monotone sequences is monotone"?
Under which circumstances the product of monotone sequences is monotone, even if it may not true for all cases? And, is there any short (general) proof without need to evaluate each single of sub-cases of the 4-cases?
Thank you.  
 A: A simple counter example to "The product of two monotone sequences is a monotone sequence" is the product of the monotone sequence $\{...,-3,-2,-1,0,1,2,3,...\}$ (which you can picture as a sequence of points in the $y$ axes of the the graph of $f(x) =x$) with itself. The product of these sequences is again a sequence viewed as a series of points in the $y$ axis of $f(x)=x^2$, and is increasing for $n>0$ but decreasing for $n<0$, as can be easily checked. So this shows that even when both sequences are increasing, their product need not be monotone. However, one can easily check that if the sequences are both increasing or both decreasing, and neither change sign, their product is monotone.
A: In general the answer is no.  Take $a_n = \left ( \frac{5}{4} \right )^n$ and $b_n = \frac{1}{n}$.  We then have 
\begin{eqnarray*}
a_1b_1 & = & \frac{5}{4} \;\; = \;\; 1.25 \\
a_2b_2 & = & \frac{25}{32} \;\; \approx\;\; 0.781 \\
a_3b_3 & = & \frac{125}{192} \;\; \approx \;\; 0.651 \\
a_{10}b_{10} & \approx & 0.93 \\
a_{15}b_{15} & \approx & 1.894.
\end{eqnarray*}
We therefore have that neither $a_nb_n \leq a_{n+1}b_{n+1}$ for all $n$, nor $a_n b_n \geq a_{n+1}b_{n+1}$ for all $n$.
