Product of bounded divergent unique numbered sequences, that product goes to a limit. I wonder if there is any example of 2 sequences $(a_n)$ and $(b_n)$, such that both of them are bounded, both of them have only unique numbers (so not $(-1)^n$ sequence), both of them have different numbers from each other and they don't converge, AND then, their product $a_i\cdot b_i$ converges to a finite limit as i approaches infinity.
Trying to find such example for a while and just cant think of any
Thanks for suggestions
 A: $$a_n=\begin{cases}\frac 1n;&2|n\\ 1-\frac 12n;&2\not|n,\end{cases}\qquad 
b_n=\begin{cases}1-\frac {\sqrt 2}n;&2|n\\ \frac {\sqrt 2}2n;&2\not|n.\end{cases}$$
A: $a_n=b_n = (-1)^n + \frac{1}{n}$
The two sequences have the properties:

*

*They are bounded,because $\forall n: -2 \leq a_n=b_n \leq 2$

*They have unique values, because
$$\begin{align}a_n=a_m&\iff (-1)^n + \frac1n = (-1)^m + \frac1m\iff  \frac1n = \frac1m \iff m=n\end{align}$$ (here, the second $\iff$ is true because the only way $(-1)^n +\frac1n = (-1)^m +\frac1m$ can be true for integer values $m,n$ is if $(-1)^m = (-1)^n$.

*Their product converges to $1$.


If you want them to have different values, i.e. if you want $a_n\neq b_n$ (though, I would like to point out that this requirement was not present in your original question) then just take $a_n = (-1)^n + \frac1n$ and $b_n = (-1)^n - \frac1n$.
A: The problem becomes fairly simple when if you realize that you can chose an an sequence you like, then choose a result sequence you like, then simply compute the necessary bn sequence. I will give an example that tries to avoid any potentially undesirable properties.
Let's define an to be sin(n)+10. That ranges between 9 and 11, it clearly does not converge, it is simple, and I don't think it has any properties anyone would consider degenerate or cheesy. For real n it repeats, which could plausibly be considered objectionable. That could easily be fixed by tweaking the function, but for integer n it is a strictly non-repeating sequence.
Let's chose to converge to 99, because it's a non-boring number and because it will give us nicer numbers later. Let's chose 1/(n^2+1) as our convergence curve - it has nice clean properties with a maximum error of 1. So our target result is 99+1/(n^2+1).
We compute bn = (99 + 1/(n^2 + 1))/(sin(n) + 10).
At infinity it ranges between 9 and 11. For non-infinite integer values of n the maximum is 11.0009 at n=6. If we consider real values on n, the true maximum is approximately 11.032 when n is approximately -1.55. In any case the function is well bounded, non-degenerate, non-converging, non-repeating, and (in my opinion) not boring or objectionable by any reasonable criteria.
