11
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Your title and initial question suggest the answer should be too short for the lower character limit on this site (at least 30 characters). If your actual question is the one in the penultimate paragraph, then please make it clear that is the (only) question you are asking. $\endgroup$ Jan 28, 2015 at 10:50
  • $\begingroup$ @ Marc van Leeuwen: the other questions are divisions of the question in title. By monotone sequence, I mean considering either increasing or decreasing for each of two, then answering to the title should include $2\times 2$ cases unless there is a general proof mentioning any cases. $\endgroup$
    – user200918
    Jan 28, 2015 at 11:58
  • 1
    $\begingroup$ Well "no" or "not always" don't require $2\times2$ cases. And in fact it is "no" or "not always" in all four cases, so distinguishing them is not terribly productive. If you are asking necessary and/or sufficient conditions for the product of monotone sequences to be monotone, then you should says so (and specify which you want). With a vague question you are likely to get not very useful answers, or none at all. $\endgroup$ Jan 28, 2015 at 13:20
  • $\begingroup$ Thank you for your informative comment. I will consider it for next(s). $\endgroup$
    – user200918
    Jan 28, 2015 at 13:39

2 Answers 2

15
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Could you please prove (not by example) the last statement of your answer? Thank you. $\endgroup$
    – user200918
    Jan 28, 2015 at 13:46
  • 2
    $\begingroup$ @Ali.E. He can't, because for monotonne 4-term sequences $(a)=1,2,3,4$ and $(b)=-40, -30, -20, -10$ the product-sequence is not monotone: $(ab)=-40, -60, -60, -40$. $\endgroup$
    – CiaPan
    Jan 28, 2015 at 13:57
  • $\begingroup$ @CiaPan: I was about to accept his answer, thank you for the rescue! Unfortunately, sometimes nobody mentioning falseness of some statements written in some answers. $\endgroup$
    – user200918
    Jan 28, 2015 at 14:02
  • $\begingroup$ @Ali.E. However, if sequences don't change sign and both are (non-)increasing or both are (non-)decreasing with respect to the absolute value, then the product-sequence will be monotone. Draw Cartesian coord system, draw some hyperbolas $xy=\mathrm{const.}$, then plot some sequences of points $(a_i,b_i)$ in each quadrant. You'll see when and why the sequence of $(a_ib_i)$ is monotone. $\endgroup$
    – CiaPan
    Jan 28, 2015 at 14:27
  • $\begingroup$ @CiaPan That's right, I assumed that both sequences had the same sign and forgot to add that. $\endgroup$
    – paoloff
    Jan 28, 2015 at 16:40
13
$\begingroup$

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 $n$.

$\endgroup$
2

You must log in to answer this question.