TITLE: If $b_n$ is a bounded sequence and $\lim a_n = 0$, show that $\lim(a_nb_n) = 0$
QUESTION [11 upvotes]: This is a real-analysis homework question so I of course have to be very precise and justify anything or any theorem I use.
If $b_n$ is a bounded sequence and $\lim(a_n) = 0$, show that $\lim(a_nb_n) = 0$
Intuitively, since $b_n$ is bounded, then sup($b_n$) is some finite number and therefore we can take an $N$ natural number as large as we need such that for all $n\gt N$ $b_na_n$ approaches $0$.
At first I thought to use the limit theorems, but since $a_n$ is not bounded, the general limit theorems do not reply. (I am referring to $\lim(X + Y) = \lim X + \lim Y$ for $X,Y$ sequences etc).
I was thinking then to use the definition of the limit somehow to show that since $b_n$ is bounded we can take as intuitively stated above $N$ large enough to show the statement is true.  I'm not sure how to proceed with this.
Thank you for your replies in advance!

REPLY [3 votes]: I think this old question deserves an answer straight from the definition of a limit.
The sequence $(b_n)$ is bounded, take an $M$ such that $\left|b_n\right|<M$ for all $n$.
Now let $\varepsilon>0$. Since $(a_n)$ converges to $0$ there is an $N$ such that $|a_n|<\frac\varepsilon M$ for all $n\ge N$. Then $|a_n b_n|<\frac\varepsilon M\cdot M=\varepsilon$ for all $n\ge N$. This shows that $(a_nb_n)$ converges to $0$.