Suppose ${a_{n}}$ and $b_{n}$ are sequence such that $\lim_{\infty }a_{n}=\lim_{\infty }b_{n}=0$ Prove that $\lim a_{n}b_{n}=0$

Question

Please check my proof

$$|a_{n}b_{n}-0|<\epsilon $$

$$ a_{n}b_{n} <\epsilon $$

$$ \frac{1}{\epsilon }< \frac{1}{a_{n}b_{n}} \leftrightarrow a_{n}b_{n}$$

Choose $a_{N}b_{N}\geq \frac{1}{\epsilon }$ for every $a_{n}b_{n}\geq a_{N}b_{N}$

there fore

$$\frac{1}{\epsilon }< \frac{1}{a_{n}b_{n}}< \frac{1}{a_{N}b_{N}}$$

the limit equal 0

Answer 1

$b_n\to 0$ implies that $|b_n|\le 1$ for all $n$ large. Accordingly, $$ |a_nb_n|=|a_n||b_n| \le |a_n| \to 0. $$

Answer 2

Maybe this answer can help:

Let $\epsilon >0$. Using the assumptions, we can find $N\in\mathbb{N}$ such that if $n\geq N$ then $|a_n|<\sqrt{\epsilon}$ and $|b_n|<\sqrt{\epsilon}$. Thus, if $n\geq N$ then $$|a_nb_n-0|=|a_n||b_n|<\sqrt{\epsilon}\cdot\sqrt{\epsilon}=\epsilon.$$ The result follows.

Answer 3

Your problem is direct application of following simple problem:

Problem: Suppose $(x_n)$ and $(y_n)$ be two sequences such that $(x_n)$ is bounded and $(y_n) \to 0$. Then $x_ny_n \to 0$ as $n \to \infty$

Solution: Use Squeeze Theorem.

Answer 4

Here is my attempt to retain the structure of the OP's proof and point out omitted details and mistakes.

Two issues with the original proof:

1. Be careful of jumping to $a_n b_n < \epsilon$ because $a_n b_n$ could be negative. For example, if $a_n b_n = -n$, then this is clearly less than any $\epsilon$ but does not converge.
2. There is subtlety in "choosing $a_Nb_N$" because these are two different sequences that might converge to $0$ at different rates. Just because $a_n < \epsilon$ for $n$ greater than some $N$ does not mean $b_n$ has reached the same level of closeness to $\epsilon.$

Fix $\epsilon > 0$. We want to show that there exists an $N$ such that $|a_nb_n - 0| < \epsilon$ for all $n \ge N.$ Notice that $$ |a_nb_n - 0| = |a_nb_n| = |a_n||b_n|. $$ Since $a_n \rightarrow 0$, there exists an $N_1$ such that $|a_n| < \sqrt{\epsilon}$ for all $n \ge N_1.$ Similarly, since $b_n \rightarrow 0$, there exists an $N_2$ such that $|b_n| < \sqrt{\epsilon}$ for all $n \ge N_2$.

Choose $N := \max{N_1,N_2}.$ Then for all $n \ge N$, $$ |a_n b_n - 0| = |a_n||b_n| < \sqrt{\epsilon}\sqrt{\epsilon} < \epsilon $$which gives our result.

Answer 5

Remarks on your proof:

Line 3 does not hold for negative $a_n b_n$. Work wih $\lvert a_n b_n\rvert$ rather.

Line 4 must focus on a choice for $N$.

Line 4 must require $a_n b_n > a_N b_N$ for the second inequality of line 6 to be true.

Suggested proof:

For any challenge $\epsilon>0$ we have $\sqrt{\epsilon}>0$ as well. As $$ \lim_{n\to\infty}a_n = \lim_{n\to\infty} b_n = 0 $$ we have $M, N \in \mathbb{N}$ such that for all $n > M$ we have $\lvert a_n\rvert < \sqrt{\epsilon}$ and for all $n > N$ we have $\lvert b_n\rvert < \sqrt{\epsilon}$.

We choose $P = \max \{ M, N \}$ then for all $n > P$ we have $$ \lvert a_n b_n \rvert = \lvert a_n\rvert \lvert b_n \rvert < \sqrt{\epsilon} \sqrt{\epsilon} = \epsilon $$ so we have $$ \lim_{n\to\infty} a_n b_n = 0 $$ as well.