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Show that (q_n)^2 approaches 2 as n goes to infinity

  1. Sep 29, 2012 #1
    For every natural number n, associate with it a natural number mn such that [itex] \frac{{m_n}^2}{n^2} < 2 < \frac{{(m_n + 1)^2}}{n^2} [/itex]. show that [itex] {q_n}^2 [/itex]approaches 2 as n approaches infinity.

    i need to show that [itex] |\frac{{m_n}^2}{n^2} - 2| < ε [/itex]. what i have so far is that [itex] 2 - \frac{{m_n}^2}{n^2} < \frac{{(m_n + 1)^2}}{n^2} - \frac{{m_n}^2}{n^2} = \frac{2m_n + 1}{n^2} [/itex].

    my trouble is that i can't figure out a way to bound m_n. i know how to take care of the n2 by choosing N such that N2ε > 1 and choosing n > N. can someone give me a hint on how to deal with the m_n? thanks.
     
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  3. Sep 29, 2012 #2

    Mute

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    Is ##q_n## supposed to be ##m_n##? If so, your original inequality can be manipulated to look like something1 < something2 - m_n^2 - 2m_n < something3, which can start you off towards a bound.

    If ##q_n## is not ##m_n##, what is it?
     
  4. Sep 29, 2012 #3
    sorry i forgot to include this: [itex] q_n = \frac{m_n}{n} [/itex]
     
  5. Sep 30, 2012 #4

    Mute

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    Ok, so your inequality is

    $$q_n^2 < 2 < \left(q_n + \frac{1}{n}\right)^2$$

    What if you try to find the limit of ##|q_n|## instead of ##q_n^2##? It looks to me like you can easily get a bound for ##|q_n|## from your inequality. The triangle inequality will come in handy for that.
     
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