Let $\{a_{n}\}\rightarrow 0$ and $\{b_{n}\}\rightarrow 0$. Prove that $\{a_{n}b_{n}\}\rightarrow 0$ using $\epsilon$-approach.

Question

Q: Let $\{a_{n}\}\rightarrow 0$ and $\{b_{n}\}\rightarrow 0$. Prove that $\{a_{n}b_{n}\}\rightarrow 0$ using $\epsilon$-approach.

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My Attempt:

Since $\{a_{n}\}\rightarrow 0$, $\exists M\in{\mathbb{R}}$ such that if $n\gt M$, then $|a_{n}-0|\lt \epsilon$ for some $\epsilon > 0$.

$\implies $if $ n>M$, $ a_{n} < \epsilon$ for some $\epsilon>0$

Similarly, $\exists M'\in{\mathbb{R}}$ such that if $n\gt M'$, then $|b_{n}-0|\lt \epsilon'$ for some $\epsilon' > 0$.

$\implies $if $ n>M'$, $ b_{n} < \epsilon'$ for some $\epsilon'>0$

Thus if $n > \max\{M,M'\}$, then $a_{n}b_{n}<\epsilon\epsilon'$

$\implies |a_{n}b_{n}-0| < \epsilon\epsilon'$ where $\epsilon\epsilon'$ is some arbitrary number $>0$

$\therefore \{a_{n}b_{n}\}\rightarrow 0$

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Did I do anything wrong or is there anything I can add to make this proof more clear? Thanks!

Answer 1

Your approach is fine.

Here is a slightly different one:

Since $b_n \to 0$ there is some $M$ such that $|b_n| \le M$ for all $n$.

let $\epsilon>0$ and choose $N$ such that for $n \ge N$ we have $|a_n| < {\epsilon \over M}$.

Then $|a_n b_n| <\epsilon$ for all $n \ge N$.

Answer 2

Your proof is correct (and also the answer by cooper.hat), but i think you should start out with a given pre-specified $\epsilon>0$, then take $n \geq \max\{M,M'\}$ such that for all $n \geq M$ we have $$|a_n | < \sqrt{\epsilon} $$ and $n\geq M'$ implies $$|b_n | < \sqrt{\epsilon}. $$ Thn we have $$|a_nb_n| <\epsilon.$$

Answer 3

More is true:

If $a_n \to 0$ and $b_n$ is bounded then $a_nb_n \to 0$.

Suppose $|b_n| < M$.

Since $a_n \to 0$, for any $c > 0$ there is a $N(c)$ such that $|a_n| < c$ for $n > N(c)$.

Then $|a_nb_n| < cM$ for $n > N(c)$.

To make $|a_nb_n| < \epsilon$, choose $cM < \epsilon$ or $c < \epsilon/M$.

Therefore, if $n > N(\epsilon/M)$, $|a_nb_n| < \epsilon$, so that $a_nb_n \to 0$.