I am trying to prove (by using the formal definition of sequence convergence) that the sequence $a_n = \left( 2 + \frac1n \right)^2$ converges to $4$
Therefore:
$$ (\forall \epsilon > 0)( \exists n_0 \in \mathbb{N})(\forall n \geq n_0): |a_n -4| <\epsilon$$
Hence,
$$ \left| \left( 2 + \frac1n \right)^2 -4 \right| <\epsilon \iff \left| 4 + \frac2n +\frac{1}{n^2}-4 \right| <\epsilon \iff$$
$$ \left| \frac{2n+1}{n^2} \right| \leq \left| 2n+1 \right| \leq \left| 2n+n \right| = \left| 3n\right| < \epsilon \iff \ $$
$$ \bbox[15px,#ffd,border:1px solid green]{n \le \frac\epsilon3}$$
Therefore in order for $a_n$ to converge at $4$ we need to choose a $n$ that satisfies the last inequality.
Is this syllogism correct?
Best Answer
The two answers posted so far are both flawed: What's needed is not a lower bound on $(4n+1)/n^2$ of the form $c/n^r$ (with $r\gt0$) but an upper bound of that form. The simplest one is
$${4n+1\over n^2}\le{5\over n}$$
(which we get by replacing the $4n+1$ with $4n+n$). From this we can see that, if $n\gt5/\epsilon$, then
$$\left|\left(2+{1\over n}\right)^2-4\right|={4n+1\over n^2}\le{5\over n}\lt\epsilon$$
Remark: I don't mean to denigrate the flawed answers or embarrass their posters (one of whom I recognize from many, many fine answers). I really only mean to point out how easy it is to make subtle mistakes when working with inequalities in epsilonish limit proofs. Everything in each answer made sense at first; it was only when I compared them that it occurred to me that if you had a choice between a lower bound of $4/n$ and $1/n^2$, then why not go all the way and say
$${1\over n^{\text{gazillion}}}\lt{4n+1\over n^2}\lt\epsilon$$
giving the gazillionth root of $1/\epsilon$ as the threshold past which you're within $\epsilon$ of the limit. But that makes no sense: No matter what you choose for $\epsilon$, the gazillionth root of $1/\epsilon$ (for sufficiently large gazillion) puts the threshold at $n=2$. I hope neither poster takes offense at my pointing out their mutual flaw. I'm quite sure I've made far more, and far more fatal, mistakes myself.