Let $\lim_{n \to \infty}a_n = a$ and $\lim_{n \to \infty}b_n = b$. We wish to prove that $$\lim_{n \to \infty}a_nb_n = ab$$

We are to prove this from the definition from convergent sequences. Let $\epsilon_1, \epsilon_2 > 0$. We have that $|a_n - a| < \epsilon_1$ and $|b_n - b| < \epsilon_2$. I have tried using some properties of absolute values to get the expression $|a_nb_n - ab| < \epsilon$, but with no success.


2 Answers 2



Use this inequality $$|a_nb_n-ab|=|a_nb_n-ab_n+ab_n-ab|\le|b_n|\cdot|a_n-a|+|a|\cdot|b_n-b|$$ and the definition of the limit of $(a_n)$ and $(b_n)$ and notice that $(b_n)$ is convergent hence bounded.

  • $\begingroup$ Ah, I seem to forget the triangle inequality. $\endgroup$ Commented Mar 19, 2014 at 18:34
  • $\begingroup$ Nice answer, Sami! $\endgroup$
    – amWhy
    Commented Mar 20, 2014 at 12:05

Consider $|a_nb_n-ab|$ = $|a_nb_n-a_nb+a_nb-ab|$ and then use triangle inequality. Then use the fact that $(a_n)$ and $(b_n)$ converge to $a$ and $b$ rpectively.

  • $\begingroup$ And the fact that convergent sequences are bounded. $\endgroup$
    – Frank
    Commented Mar 19, 2014 at 18:32
  • $\begingroup$ yeah right...i forgot to mention that. $\endgroup$
    – wanderer
    Commented Mar 19, 2014 at 18:36

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