# Give an example such that $(x_ny_n)$ converges but $(x_n)$ and $(y_n)$ diverges.

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.

• 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). – Clarinetist Jan 9 '18 at 21:06
• Looks good to me! You could have $x_n$ and $y_n$ alternate between 0 and 1, but offset so they multiply to 0. – Cameron Williams Jan 9 '18 at 21:07
• Thanks for comments. – PozcuKushimotoStreet Jan 9 '18 at 21:10
• 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 – 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.
• The question asks about $a_n b_n$ – Mark Jan 9 '18 at 21:15