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 $4$


$$ (\forall \epsilon > 0)( \exists n_0 \in \mathbb{N})(\forall n \geq n_0): |a_n -4| <\epsilon$$


$$ \left| \left( 2 + \frac1n \right)^2 -4 \right| <\epsilon \iff \left| 4 + \frac2n +\frac{1}{n^2}-4 \right| <\epsilon \iff$$

$$ \left| \frac{2n+1}{n^2} \right| \leq \left| 2n+1 \right| \leq \left| 2n+n \right| = \left| 3n\right| < \epsilon \iff \ $$

$$ \bbox[15px,#ffd,border:1px solid green]{n \le \frac\epsilon3}$$

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?


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.


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

A correct way is to use a lower bound


and the convergence condition is established for all



Shame on me, this was plain wrong.

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

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

  • $\begingroup$ 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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