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$a_n, b_n$ are bounded sequences. If $a_n$ converges to zero and $b_n$ diverges, then $a_nb_n$ converges.

I think that the fact that if one was divergent and one convergent, this fact would be false, but does the boundedness change the outcome to be a true fact?

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    $\begingroup$ You can use the bounds on $b_n$ to bound the product sequence. Specifically, $a_nb_{min} \leq a_nb_n \leq a_n b_{max}$,and the bounding sequences easily converge. $\endgroup$
    – Paul
    Mar 27, 2020 at 19:44
  • $\begingroup$ Where is a “sum” of sequences in your question? Perhaps you meant “product”? $\endgroup$
    – Martin R
    Mar 27, 2020 at 21:02

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Hint: The fact that $a_n\to0$ and that $(b_n)$ is bounded implies that $|a_nb_n|\leq|a\cdot M|=M|a|$ for some constant $M>0$, if you go far enough out in your sequence. Hence, what can you say about the convergence of $a_nb_n$, given that you know $a_n\to0$?

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  • $\begingroup$ would $a_nb_n$ then be bounded by M? $\endgroup$
    – Xergonuno
    Mar 27, 2020 at 19:51
  • $\begingroup$ I mean be bounded by 0 $\endgroup$
    – Xergonuno
    Mar 27, 2020 at 19:51
  • $\begingroup$ Bounded by $0$ would imply that the sequence $a_nb_n$ is a constant sequence. You don't need anything specific about the boundedness of $a_n$ to conclude (although $a_n$ is bounded, since it is convergent), just that it converges to $0$. Think about the $\varepsilon-N$ definition of convergence. $\endgroup$
    – csch2
    Mar 27, 2020 at 19:53
  • $\begingroup$ so then $|a_n| < \epsilon$ and $|b_n| <= M$ so then could you say $|a_n| < \frac{epsilon - L}{M}$? $\endgroup$
    – Xergonuno
    Mar 27, 2020 at 19:58
  • $\begingroup$ Because then $|a_nb_n - L| <= |a_n||b_n| + L = (\frac{\epsilon - L}{M} \dot M) + L = \epsilon$ $\endgroup$
    – Xergonuno
    Mar 27, 2020 at 19:59
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We have $\;\lim\limits_{n\to\infty}a_n=0\;,\;\;|b_n|\le M\;$ , so applying the squeeze theorem + arithmetic of limits:

$$0\xleftarrow[\infty\leftarrow n]{}-Ma_n\le a_nb_n\le a_nM\xrightarrow[n\to\infty]{}0$$

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1) $|b_n| <M >0$, since $b_n$ is bounded.

2) $0\le |a_nb_n| <M |a_n|$.

3) $a_n$ converges to $0$, then $|a_n|$ converges to $0$.

4) Let $\epsilon/M$ be given. There is a $n_0$ s.t.

$n\ge n_0$ implies $|a_n| <\epsilon/M$.

Then

$|a_nb_n|<M|a_n| <M(\epsilon/M) <\epsilon$.

5) $|a_nb_n|$ converges to $0$, then $a_nb_n$ converges to $0$.

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