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I am trying to find example of two bounded monotone sequences, one increasing and one decreasing, such that the product is not a monotone sequence.

I have tried many of examples, but all my products are monotone. It makes me wonder if the boundedness somehow makes the sequences end up always monotone? Of course, it is easy to find non-bounded monotone functions such that the product is not monotone.

Thanks for the help, sorry if bad English.

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There's a trivial-ish example: $(-1, -1, 1, 1, 1, \dots)$ and $(1, 0, -1, -1, -1, \dots)$, where the product is $(-1, 0, -1, -1, \dots)$.

I don't know if it's possible if we use "eventually monotone" instead of "monotone"; certainly the two sequences must be convergent, and we can't produce an oscillation around $0$.

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Probably not the simplest example

Suppose your first sequence is $a_n = 1+ \dfrac{3+(-1)^n}{3^n}$ so $\frac{5}{3},\frac{13}{9},\frac{29}{27},\frac{85}{81},\frac{245}{243},\frac{733}{729},\ldots$ so monotonic decreasing and bounded by $1$ and $2$

and your second sequence is $b_n=\dfrac{1}{a_{n+1}}$ so monotonic increasing and bounded by $\frac12$ and $1$

so their product is $ \dfrac{3^{n+1}+9+3(-1)^n}{3^{n+1}+ 3-(-1)^{n}}$ so $\frac{15}{13},\frac{39}{29},\frac{87}{85},\frac{255}{245},\frac{735}{733},\frac{2199}{2189},\ldots$ which is not monotonic: terms for even $n$ are bigger than their neighbours

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My try:

Let $$a_n=\begin{cases} 2-\frac{1}{2^n} & \text{for odd n} \\ 2-\frac{1}{2^n}-\frac{1}{10^n} &\text{for even n} \end{cases}$$ $$b_n=\begin{cases} (2-\frac{1}{2^n}-\frac{1}{10^n})^{-1} & \text{for odd n} \\ (2-\frac{1}{2^n})^{-1} &\text{for even n} \end{cases}$$

They are both monotonic.

For odd n, $a_n b_n>1$. For even n, $a_n b_n<1$.

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