I intuitively understand proof with limits, but I'm not sure on how to write a formal proof for this example.

For each $n \in \mathbb{N}$, let $a_n$, $b_n$ be real numbers. Also, let $a_{\infty}$, $b_{\infty}$ be real numbers. Suppose that $\lim_{n\to \infty} a_n = a_{\infty}$ and $\lim_{n\to \infty} b_{n} = b_{\infty}$.

Prove that $\lim_{n\to \infty} (a_n b_n) = a_\infty b_\infty$.

  • $\begingroup$ Do you mean $\displaystyle\lim_{n \to \infty}(a_nb_n) = a_{\infty}b_{\infty}$ or $\displaystyle\lim_{n \to \infty}(a_n+b_n) = a_{\infty}+b_{\infty}$? $\endgroup$ – JimmyK4542 Jul 24 '14 at 22:51
  • $\begingroup$ I meant the former one. I changed it. $\endgroup$ – user164179 Jul 24 '14 at 22:54
  • $\begingroup$ Scroll down to "Proof of the Product Rule for Limits" en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules $\endgroup$ – JimmyK4542 Jul 24 '14 at 22:56
  • 1
    $\begingroup$ $a_nb_n-a_{\infty}b_{\infty}=a_n(b_n-b_{\infty})+(a_n-a_{\infty})b_{\infty}$ $\endgroup$ – Hamou Jul 24 '14 at 22:56

You want to show that $\lim_{n \to \infty} a_nb_n = ab$, where $\lim_{n \to \infty} a_n = a$ and similarly $\lim_{n \to \infty} b_n = b$. By the Triangle Inequality, we have that

\begin{align} |a_nb_n - ab| &= |(a_nb_n - a_nb) + (a_nb - ab)| \\ &\leq |a_n(b_n - b)| + |b(a_n - a)| \\ &= |a_n||b_n - b| + |b||a_n - a|. \end{align}

Since $\lim_{n \to \infty} a_n = a$, given $\epsilon > 0$, there exists $M_1 > 0$ such that $|a_n| < M_1$ for all $n \in \mathbb{N}$. Set $M = \sup\{M_1,b\}$. Then we have

$$|a_nb_n - ab| \leq M|b_n - b| + M|a_n - a|.$$

Now from the convergence of the sequences $(a_n)$ and $(b_n)$, we have that if $\epsilon > 0$ is given, there exist $K_1,K_2 \in \mathbb{N}$ such that if $n \geq K_1$, $|a_n - a| < \epsilon/2M$ and analogously, if $n \geq K_2$ then $|b_n - b| < \epsilon/2M$. Set $K = \sup\{K_1,K_2\}$.

What can you conclude on the quantity $|a_nb_n - ab|$ if $n \geq K$?


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