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If $b_n$ is a bounded sequence and $\lim (a_n)= 0$ , show that $\lim(a_nb_n)=0$ and explain why the next Theorem cannot be used.

Theorem: If $x_n$ converges to $x$ and $y_n$ converges to $y$ then $\lim(x_ny_n)=xy.$


To use the theorem both sequences should be convergents. Sequence $b_n$ is a bounded sequence but may not be convergent.


Proof

I know $|x_n|<\varepsilon, \varepsilon>0 $ and by definition of bounded sequence $|b_n|\leq B, B>0$ . How can I continue with my proof on this?

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    $\begingroup$ In your "I know" section you forget about to say "for all" or "there exists". These terms are very important. $\endgroup$ – amsmath Nov 5 '17 at 16:28
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We know $\forall \varepsilon >0, \exists N \in \mathbb{N}_+$ such that $|a_n| < \frac{\varepsilon}{M}$ for all $n>N$.

We also know that $b_n$ is bounded, so $|b_n|<M$ for all $n \in \mathbb{N}_+$. Now consider $$|a_n b_n| = |a_n| |b_n|<\varepsilon$$ for all $n>N$as required.

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Since $(b_n)_{n=1}^\infty$ is bounded, there exists $M > 0$ such that $|b_n| \le M$ for all $n\in\mathbb{N}$.

We have:

$$0 \le |a_nb_n| \le M |a_n|\xrightarrow{n\to\infty} 0$$

So by the squeeze theorem we obtain $\lim_{n\to\infty} |a_nb_n| = 0$, which is equivalent to $\lim_{n\to\infty} a_nb_n = 0$.

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The theorem that you indicated cannot be used simply because $y_{n}$ is not convergent. For the solution, let $M$ be the bound, then $|a_{n}b_{n}|\leq M|a_{n}|$, one can use squeeze theorem to deduce the result.

As a request, I will prove that $M|a_{n}|\rightarrow 0$. Given $\epsilon>0$, since $a_{n}\rightarrow 0$, find some $N\in{\bf{N}}$ such that for $n\geq N$, $|a_{n}|<\epsilon/(M+1)$, then $M|a_{n}|<\epsilon$.

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  • $\begingroup$ What if OP doesn't know that $Ma_n\to 0$? $\endgroup$ – amsmath Nov 5 '17 at 16:33
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    $\begingroup$ That would be ridiculous @amsmath $\endgroup$ – Dionel Jaime Nov 5 '17 at 16:34
  • $\begingroup$ Why would it be ridiculous? Because it's trivial? Well, the whole question is trivial (for us). $\endgroup$ – amsmath Nov 5 '17 at 16:36

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