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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

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    $\begingroup$ Your proof makes no sense to me. You should start with the words "Suppose $\epsilon$ is given to me (probably a pretty small number). Then I can find an integer $N$ (which may have to be large) such that ..." $\endgroup$ Dec 24, 2016 at 13:27
  • $\begingroup$ @EthanBolker Thank you.:) $\endgroup$
    – Lingnoi401
    Dec 24, 2016 at 13:29
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    $\begingroup$ These manipulations aren't correct when $a_nb_n$ is negative, but that's easily fixed by considering $|a_nb_n|$ instead. $\endgroup$
    – Sophie
    Dec 24, 2016 at 13:30
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    $\begingroup$ @Sophie I think the real problem with the "proof" is the way it's written. Once the OP gets that right then fixing the problem you (correctly) point out is easy. The logic is more important than the manipulations. $\endgroup$ Dec 24, 2016 at 13:33
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    $\begingroup$ It would be useful to note that $x_n \to 0$ if and only if $|x_n|\to 0$, which follows directly from the definition of limit.. $\endgroup$ Dec 24, 2016 at 13:33

5 Answers 5

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

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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.

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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.

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  • $\begingroup$ This is a correct argument - but I think the OPs real problem is not how to prove the result, but how to structure an $\epsilon-\delta$ or $\epsilon-N$ proof. $\endgroup$ Dec 24, 2016 at 13:50
  • $\begingroup$ Dear @EthanBolker: Since that was already mentioned in other answers so i gave a 'different' solution! $\endgroup$ Dec 24, 2016 at 13:51
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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.

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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.

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