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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?

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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.

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No. What you are doing is starting from $A<\epsilon$, noticing that $A<B$ and concluding $B<\epsilon$. This is not a valid reasoning.

What you can do instead is $$\epsilon>\left|\frac{4n+1}{n^2}\right|>\frac{1}{n^2}$$ giving $n>\tfrac{1}{\sqrt\epsilon}$. Notice that this makes more intuitive sense, as $n$ needst to get bigger if you want to have a smaller $\epsilon$.

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    $\begingroup$ @Veriun Yes, but that is not what you did. Check your steps carefully, it's very important that you understand your mistake in this case. $\endgroup$ Jul 27 '20 at 7:33
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    $\begingroup$ @Veriun You're welcome :) $\endgroup$ Jul 27 '20 at 7:34
  • $\begingroup$ By the way, shouldn't it be $\epsilon > \frac{1}{n^2} \iff n^2 > \frac1\epsilon \iff n > \frac{1}{\sqrt \epsilon} = \epsilon^{-\frac12}$? $\endgroup$
    – Dimitris
    Jul 27 '20 at 8:25
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    $\begingroup$ @Veriun Right, thanks. Edited. $\endgroup$ Jul 27 '20 at 8:27
  • $\begingroup$ The OP made a minor error: It should be $4n+1$, not $2n+1$. $\endgroup$ Jul 27 '20 at 13:13
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A correct way is to use a lower bound

$$\frac4n<\frac{4n+1}{n^2}<\epsilon$$

and the convergence condition is established for all

$$n>\frac4\epsilon.$$


Update:

Shame on me, this was plain wrong.

You need an upper bound which is still lower than $\epsilon$ (thanks Barry).

$$n\ge\frac 5\epsilon\implies\frac{4n+1}{n^2}<\frac 5n<\epsilon.$$

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  • $\begingroup$ It should be 4n+1, not 2n+1. And $\epsilon\gt2/n$ does not imply $\epsilon\gt(2n+1)/n^2$ $\endgroup$
    – Empy2
    Jul 27 '20 at 9:49
  • $\begingroup$ @Empy2: $2$ fixed. The implication is the other way. $\endgroup$
    – user65203
    Jul 27 '20 at 10:28
  • $\begingroup$ Please see my answer, just now posted. $\endgroup$ Jul 27 '20 at 14:16
  • $\begingroup$ @BarryCipra: yep, I shoot myself a bullet in the foot ! $\endgroup$
    – user65203
    Jul 27 '20 at 14:32

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