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The sequence $a_n = \dfrac{n^2-1}{n^2+n+1}$ is said to converge to 1 . When we divide the numerator with the denominator we get $$a_n = 1-\dfrac{(n+2)}{n^2+n+1}$$. The text I am referring to says, we only need to show that, $$r_n = \dfrac{n+2}{n^2+n+1}$$ converges to 0 as n tends to infinity and proceeds to say that $$\forall n>2, n+2<2n$$ and that $$n^2+n+1>n^2$$. I get the proof till this point, then the author states $$0<r_n<2n/n^2=2/n$$ for n>2, and says this determines the largest estimate of how much the sequence $a_n$ differs from 1. What did the author do here? why and how did they bound $r_n$? How does this (how does bounding $r_n$ ..?) prove that the limit of $a_n$is infact 1

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    $\begingroup$ I think the author intended to define $r_n$ as $\frac{n + 2}{n^2 + n + 1}$, and then show it tends to $0$, not $1$. The squeeze argument in the final line shows $r_n \to 0$. $\endgroup$ Commented Sep 11, 2023 at 17:48
  • $\begingroup$ Oh typo, Let me edit it $\endgroup$ Commented Sep 11, 2023 at 17:48
  • $\begingroup$ By tend to 1 , i meant $a_n$ tends to 1 sorry for the misunderstanding $\endgroup$ Commented Sep 11, 2023 at 17:49
  • $\begingroup$ I'll make it more clear $\endgroup$ Commented Sep 11, 2023 at 17:50
  • $\begingroup$ All done , my bad ^⁠_⁠^ $\endgroup$ Commented Sep 11, 2023 at 17:51

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Observe and be sure you understand and can prove the following inequalities:

$$0\le r_n=\frac{n+2}{n^2+n+1}\le \frac{n+n}{n^2}=\frac 2n\xrightarrow[n\to\infty]{}0$$

...and now apply the squeeze theorem.

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