# Proof that if $\lim_\limits{n\to\infty}(a_n\cdot b_n)=0$ then $\lim_\limits{n\to\infty}a_n$ or $\lim_\limits{n\to\infty}b_n = 0$

We had to prove that if

$$\lim_{n\to\infty}(a_n\cdot b_n)=0$$

Then either $\lim_{n\to\infty}a_n$ or $\lim_{n\to\infty}b_n$ HAS to be equal to $0$.

My hypothesis is that since

$$\lim_{n\to\infty}(a_n\cdot b_n)=\lim_{n\to\infty}a_n\cdot \lim_{n\to\infty}b_n$$

Then for $\lim_{n\to\infty}(a_n\cdot b_n)$ to be zero, and since the only "number" that when multiplied by another one produces $0$ (or something along those lines), at least one of the factors ($a_n$ and $b_n$) MUST be $0$.

But the thing is that we couldn't come up with any formal proof, using the definition of limit or something... So any advice would be appreciated.

Thanks.

• If in the limit neither is zero, the product will be non-zero. What kind of formalism are you looking for? Commented Nov 13, 2016 at 21:19
• @darksky he forgot to assume that at least one of the sequences converges. Commented Nov 13, 2016 at 21:19
• In the problem we don't assume that at least of the sequences converges, we have to prove or disprove that one of them converges to zero. But someone already gave a counterexample. Commented Nov 13, 2016 at 21:41

The claim is false. Consider $$a_n=\begin{cases}0, & n\:\:\mathrm{even}, \\1, & n\:\:\mathrm{odd}, \end{cases}$$ and $$b_n=\begin{cases}1, & n\:\:\mathrm{even}, \\0, & n\:\:\mathrm{odd}. \end{cases}$$
We have $a_nb_n=0,\forall n$ and thus $\lim_{n\infty} a_nb_n=0.$ But $\lim_{n\to\infty}a_n$ and $\lim_{n\to\infty}b_n$ don't exist.
Take: $a_n=\left\{\begin{matrix} 1/n^2 & n \: odd \\ n & n \: even \end{matrix}\right., \; \; b_n=\left\{\begin{matrix} 1/n^2 & n \: even \\ n & n \: odd \end{matrix}\right.$
Then $a_n\cdot b_n = 1/n$ for every $n$ but non of the sequences has a limit.