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I'm trying to prove this proposition:

Theorem: Suppose $(a_n)$ is a sequence that converges to $0$, and that $(b_n)$ is a bounded sequence. Then the sequence $(a_n \cdot b_n)$ is convergent and $$ \lim_{n \to \infty} a_n \cdot b_n = 0. $$

Attempt at proof: Because the sequence $(b_n)$ is bounded, there exists an $M > 0$ such that $|b_n| \leq M$ for every $n \in \mathbb{N}$. Then we have $$ 0 \leq | a_n \cdot b_n | \leq M |a_n |. $$ Because $(a_n)$ converges to zero, there exists a $n_0 \in \mathbb{N}$ such that $|a_n| < \epsilon$ for all $n \geq n_0$. Since $M > 0$ we thus also have $$ 0 \leq | a_n \cdot b_n | \leq M |a_n | < M \epsilon. $$

Now I'm stuck and don't know how to proceed. Any help would be appreciated.

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  • $\begingroup$ You are correct. $\endgroup$ – sinbadh Jan 20 '16 at 19:07
  • $\begingroup$ Just swap $\epsilon$ for $\frac{\epsilon}{M}$ $\endgroup$ – Anthony Peter Jan 20 '16 at 19:13
  • $\begingroup$ Let $n>n_0$ imply that $|a_n|<\frac{\epsilon}{M}$. Then, for all $\epsilon>0$, $|a_nb_n|\le M|a_n|<M\frac{\epsilon}{M}=\epsilon$ $\endgroup$ – Mark Viola Jan 20 '16 at 19:13
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You are done. A way to see it better: for any $\varepsilon$ you can find $n_0$ such that $|a_n|<\frac{\varepsilon}{M}$ for $n > n_0$ and so $$0 \leq |a_n \cdot b_n| \leq M |a_n| < M \frac{\varepsilon}{M} = \varepsilon,$$ so $a_n \cdot b_n \to 0$.

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