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

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

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Mute
Homework Helper
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?

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