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

  • #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.
 

Answers and Replies

  • #2
Mute
Homework Helper
1,388
10
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?
 
  • #3
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?
sorry i forgot to include this: [itex] q_n = \frac{m_n}{n} [/itex]
 
  • #4
Mute
Homework Helper
1,388
10
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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