Give an example of sequences $(x_n)$ and $(y_n)$ in $\mathbb{R}$ such that $(x_ny_n)$ converges but $(x_n)$ and $(y_n)$ diverges.

My answer: Let $x_n=(-1)^n$ and $y_n=(-1)^n$. Then, $x_ny_n=(-1)^n(-1)^n=(-1)^{2n}$. So, $(-1)^{2n}$ convergence to $1$ and $(x_n)$, $(y_n)$ diverges.

Can you check my answer?

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    $\begingroup$ Your example looks fine to me; however, you may want to prove (if you haven't already) that $\left((-1)^n\right)_{n \in \mathbb{N}}$ is divergent and $(1)_{n \in \mathbb{N}}$ is convergent (the latter is trivial). $\endgroup$ – Clarinetist Jan 9 '18 at 21:06
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    $\begingroup$ Looks good to me! You could have $x_n$ and $y_n$ alternate between 0 and 1, but offset so they multiply to 0. $\endgroup$ – Cameron Williams Jan 9 '18 at 21:07
  • $\begingroup$ Thanks for comments. $\endgroup$ – PozcuKushimotoStreet Jan 9 '18 at 21:10
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    $\begingroup$ Your answer is not only correct, but it is probably one of the easiest and most forward of all examples that one can come up with. Nice. +1 $\endgroup$ – DonAntonio Jan 9 '18 at 21:31

We can take $$a_n=(2+(-1)^n) $$ $$b_n=(2-(-1)^n) $$ $$a_nb_n=3$$

or $$a_n=\sin (n) \;\;,\; \;b_n=\frac {1}{\sin (n)} $$


Your answer is correct. $$ a_n = sin(n+1)$$ and $$b_n = csc (n+1)$$ where $ {a_n}{b_n}$ converges to $1$ while both $ a_n$ and $ b_n$ diverge.

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    $\begingroup$ The question asks about $a_n b_n$ $\endgroup$ – Mark Jan 9 '18 at 21:15

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