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

[demonelite123]

For every natural number n, associate with it a natural number m_n such that \frac{{m_n}^2}{n^2} < 2 < \frac{{(m_n + 1)^2}}{n^2}. show that {q_n}^2approaches 2 as n approaches infinity.

i need to show that |\frac{{m_n}^2}{n^2} - 2| < ε. what i have so far is that 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}.

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.

 

[Mute]

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?

 

[demonelite123]

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: q_n = \frac{m_n}{n}

 

[Mute]

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.