# Sum of divergent and convergent sequences

$$a_n, b_n$$ 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?

• 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.
– Paul
Mar 27, 2020 at 19:44
• Where is a “sum” of sequences in your question? Perhaps you meant “product”? Mar 27, 2020 at 21:02

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 $$a_n\to0$$?

• would $a_nb_n$ then be bounded by M? Mar 27, 2020 at 19:51
• I mean be bounded by 0 Mar 27, 2020 at 19:51
• 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. Mar 27, 2020 at 19:53
• so then $|a_n| < \epsilon$ and $|b_n| <= M$ so then could you say $|a_n| < \frac{epsilon - L}{M}$? Mar 27, 2020 at 19:58
• Because then $|a_nb_n - L| <= |a_n||b_n| + L = (\frac{\epsilon - L}{M} \dot M) + L = \epsilon$ Mar 27, 2020 at 19:59

We have $$\;\lim\limits_{n\to\infty}a_n=0\;,\;\;|b_n|\le M\;$$ , so applying the squeeze theorem + arithmetic of limits:

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

1) $$|b_n| 0$$, since $$b_n$$ is bounded.

2) $$0\le |a_nb_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|.

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