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Let $(a_n)$ and $(b_n)$ be positive real sequences such that $\lim \limits_{n \to \infty} \dfrac{a_n}{b_n} = 0$ and $(b_n)$ is bounded. Prove that $\lim \limits_{n \to \infty} a_n=0$.

Proof: Since $(b_n)$ is bounded it converges, say $\lim b_n = L$. Given $\varepsilon > 0$, then there exists $n_1,n_2 \in \mathbb{N}$ such that $|a_n - L| < \varepsilon$ for all $n>n_1$ and $|b_n-L| < \varepsilon$ for all $n>n_2 $. I Don't really know where to go from here...any hint on how to approach this?

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  • $\begingroup$ $b_n$ doesn't have to converge just because it's bounded. The sequence $|\sin(n)|$ is bounded by $1$, but it doesn't converge to anything. $\endgroup$ – Arthur Oct 22 '14 at 0:54
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You don't have enough info to infer that $\{b_n\}$ converges.

Let $M>0$ be an upperbound of $\{b_n\}$. For a given $\epsilon>0$, there is $N>0$ such that for $n\geq N$, we have $|a_n/b_n|<\frac{\epsilon}{M}$. Then, for all such $n$, $$ \frac{\epsilon}{M}>|a_n/b_n|=a_n/b_n\geq a_n/M\implies0<a_n<\epsilon\implies|a_n|<\epsilon. $$

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If $a_n/b_n \to 0$, then, for any $c > 0$, there is an $N(c)$ such that $|a_n/b_n| < c$ for $n > N(c)$, or $|a_n| < c|b_n|$.

If $|b_n|$ is bounded, there is a $B > 0$ such that $|b_n| < B$.

Therefore $|a_n| < cB$ for $n > N(c)$.

Finally, choose $c = \epsilon/B$. Then $|a_n| < \epsilon $ for $n > N(\epsilon/B)$.

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