# Multiply a Known Convergent Sequence by a Bounded Sequence

I have been tasked with proving the following:

$$a_n$$ is bounded but not necessarily convergent

assume that $$\lim{b_n} \to 0$$

show $$\lim{a_n b_n} \to 0$$.

I started my proof listing what I know:

1. If $$a_n$$ is bounded then there is some number $$M \gt 0$$ such that $$|a_n| \le M$$ for all $$n \in \mathbb{N}$$.
2. $$|b_n - 0| \lt \epsilon$$.

My proof proceeded directly from this:

$$|a_n b_n| = |a_n| |b_n| \lt \epsilon$$

$$|b_n| \lt \frac{\epsilon}{|a_n|}$$

And since $$|b_n - 0| = |b_n|$$ can be made arbitrarily small by (2) above we have shown that for all $$n \ge N$$ ($$N \in \mathbb{N}$$),

$$|a_n b_n| = |a_n| |b_n| \lt |a_n| \frac{\epsilon}{|a_n|} = \epsilon$$.

Our limit therefore converges to 0 as desired.

I was fairly satisfied with this proof, I had considered $$M$$ being able to be replaced by $$|a_n|$$ (as I didn't see a reason I couldnt do this), so this took advantage of (1) as well.

The author went about it in a different but similar way starting from the same "base":

$$|a_n b_n - 0| = |a_n| |b_n| \le M|b_n|$$ from (1)

Because $$(b_n) \to 0$$ we can pick an $$N$$ such that:

$$|b_N| \lt \frac{\epsilon}{K}$$

Finally, we can conclude that for this choice of $$N$$,

$$|a_n b_n - 0| \lt K |b_n| \lt K \frac{\epsilon}{K} = \epsilon$$

The author seemed to have just "dropped" the $$|a_n|$$ from the last part where he does $$K |b_n|$$?

Other than that weirdness (and the ham fisted nature of my newbie proof), is there any fundamental differences in our approach? Am I not able to let $$M = |a_n|$$?

Your proof is fine, but it's hard to read. Here's a revised version:

Let $$M$$ be a positive number such that $$|a_{n}|\leq M$$ for all $$n$$. Then, $$\left|a_{n}b_{n}\right| =\left|a_{n}\right|\left|b_{n}\right| =M\left|b_{n}\right|.$$ For each $$\epsilon>0$$, pick $$N$$ large enough such that $$|b_{n}|<\epsilon/M$$ whenever $$n\geq N$$. It follows that $$|a_{n}b_{n}| whenever $$n\geq N$$. Therefore, $$a_{n}b_{n}\rightarrow0$$.

• Small typo $\leq$ in middle equation. Feb 16, 2019 at 3:32
• @JoaquinSan: +1 thanks, fixed! Technically not wrong... ;-) Feb 16, 2019 at 3:33
• Thank you for your answer. An answer below says my proof is wrong due to the dependence of n in my proof in the epsilon half of the inequality. If you could address this for me and why it's right or wrong (relative to your answer here) I will happily mark this answered.
– CL40
Feb 16, 2019 at 8:16
• It's because you say let $M=|a_n|$ so it depends on $n$. You can fix it by taking $M=\sup_n|a_n|$. Feb 16, 2019 at 21:08
• I meant the second one, not the first. $M$ is an upper bound not necessarily on of the $a_n$. Feb 17, 2019 at 6:46

Your mistake is when you put $$\dfrac{\epsilon}{\vert{a_n} \vert}$$ this expression depends on $$n$$ so your inequality is only valid for that $$n$$.

When you use the fact that $$\vert a_n \vert \leq M, \forall n \in \mathbb{N}$$ then that number $$M$$ can be used at your convenience to find values ​​that meet the definition of limit.

What you want to have is that $$\vert a_n b_n \vert < \epsilon$$ for very large values ​​of $$n$$. So if you give some $$\epsilon >0$$ the value $$\dfrac{\epsilon}{M} >0$$ and applying the definition of limit you should that : $$\exists$$ $$n_0 \in \mathbb{N} / n>n_0 : \vert b_n \vert < \dfrac{\epsilon}{M}$$.

Now for $$n>n_o : \vert a_n b_n \vert = \vert a_n \vert \vert b_n \vert < M. \dfrac{\epsilon}{M} = \epsilon$$ and you have what you were looking for.