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If $(x_n)$ and $(y_n)$ are positive real sequences such that $(y_n)$ is bounded and $\lim\frac{(x_n)}{(y_n)} = 0$, prove that $\lim(x_n) = 0$

So since $(y_n)$ is bounded, there exists a real number $M$ such that for all n natural numbers, $(y_n) < |M|$ So by the limit definition :
$\lim|\frac{(x_n)}{(y_n)}-0| = \lim\frac{(x_n)}{(y_n)} < \epsilon$ So $|(x_n)| < \epsilon(y_n)$

Now I need to find a natural number $N$ such that for all $n > N$, $|(x_n)| < \epsilon$ is this correct which would prove this statement. However I'm having trouble picking something that will work here? Am I on the right track? I have to be very careful also about anything I use to quote the theorem since this is an introduction to real analysis class.

Thanks!

2 Answers 2

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I would approach this by contradiction:

If the sequence does not converge to 0, then there is a subsequence that is bounded away from 0. So we may fix $a>0$ such that $|x_n|>a$ for infinitely many values of $n$.

Since the sequence of $y_n$ is bounded, say by $M$, we have $|y_n| for all $n$, so if $n$ is one of the indices in our subsequence, we have $|x_n/y_n|>a/M$.

This inequality holds for infinitely many $n$, and we have found a subsequence of $(x_n/y_n)$ that does not converge to 0.

Does this make sense?

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    I'm just not seeing how$a$subsequence is relating to every case for $\frac{|x_n|}{|y_n|}$ since we need to show that $\frac{|x_n|}{|y_n|}$ > something for ALL n?2010-12-01
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Hint

If $|y_{n}|\leq M$ then $1 \leq \frac{M}{|y_{n}|} $

$| x_{n}| \leq |x_{n}| \cdot \frac{M}{|y_{n}|} = |\frac{x_{n}}{y_{n}}| \cdot M < \varepsilon_{0} \cdot M = \frac{\varepsilon}{M} \cdot M = \varepsilon$

Where $|\frac{x_{n}}{y_{n}}| < \varepsilon_{0}$ *for hypotesis* and choosing $\varepsilon_{0} = \frac{\varepsilon}{M}$

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    @Hans Lundmark: Thanks for the reponse.2010-11-30