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 $M = |a_n|$?


2 Answers 2


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 $|a_{n}b_{n}|<M(\epsilon/M)=\epsilon$ whenever $n\geq N$. Therefore, $a_{n}b_{n}\rightarrow0$.

  • 1
    $\begingroup$ Small typo $\leq$ in middle equation. $\endgroup$ Feb 16, 2019 at 3:32
  • $\begingroup$ @JoaquinSan: +1 thanks, fixed! Technically not wrong... ;-) $\endgroup$
    – parsiad
    Feb 16, 2019 at 3:33
  • $\begingroup$ 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. $\endgroup$
    – CL40
    Feb 16, 2019 at 8:16
  • $\begingroup$ It's because you say let $M=|a_n|$ so it depends on $n$. You can fix it by taking $M=\sup_n|a_n|$. $\endgroup$
    – parsiad
    Feb 16, 2019 at 21:08
  • $\begingroup$ I meant the second one, not the first. $M$ is an upper bound not necessarily on of the $a_n$. $\endgroup$ 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.


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