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Let $(a_n)$ and $(b_n)$ be two sequences with $$ \lim_{n\to\infty}\frac{a_n}{b_n}=1, $$ which one usually writes as $a_n\sim b_n$. Let $b_n\to 0$ as $n\to\infty$. My question is, if then $a_n\to 0$, too.

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    $\begingroup$ From the limit condition, eventually $|a_n|\le2|b_n|$. $\endgroup$ Jan 2, 2015 at 11:44

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We have $a_n\sim b_n$ so there's a sequence $(\epsilon_n)$ convergent to $0$ such that

$$a_n=(1+\epsilon_n)b_n$$ and the result follows easily.

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Hint: assume the opposite. If $a_n\to 0$ were wrong, then we would have $|a_n|>\varepsilon$ for arbitrarily large $n$. However, since $b_n\to 0$, what would that tell us about $|\frac{a_n}{b_n}|$?

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  • $\begingroup$ Then $\left\lvert\frac{a_n}{b_n}\right\rvert$ would diverge, but by assumption it is $\left\lvert\frac{a_n}{b_n}\right\rvert\to 1$. Right? $\endgroup$
    – user34632
    Jan 2, 2015 at 11:49
  • $\begingroup$ But it's actually not true, is it? $\endgroup$
    – servabat
    Jan 2, 2015 at 11:49
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    $\begingroup$ @servabat What exactly do you think is wrong here? $\endgroup$ Jan 2, 2015 at 11:53
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    $\begingroup$ @math12 You are correct. The conclusion contradicts one of your assumptions. You know that $a_n\sim b_n$ and that $b_n\to0$, so the only assumption that could possibly be false is $a_n\not\to0$. Therefore, $a_n\to0$. $\endgroup$
    – Regret
    Jan 2, 2015 at 11:56
  • $\begingroup$ Well, I just remember some exceptions when the limit was 0 (it was working only for a limit different from 0) but I'm probably confused then $\endgroup$
    – servabat
    Jan 2, 2015 at 12:02
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We have $\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}\left(\frac{a_n}{b_n}\cdot b_n\right)=\left(\lim_{n\to\infty}\frac{a_n}{b_n}\right)\left(\lim_{n\to\infty}b_n\right)=1\cdot0=0$.

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