# Question about convergence of a sequence

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 $$0 for n>2, and says this determines the largest estimate of how much the sequence $$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

• 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 $r_n \to 0$. Commented Sep 11, 2023 at 17:48
• Oh typo, Let me edit it Commented Sep 11, 2023 at 17:48
• By tend to 1 , i meant $a_n$ tends to 1 sorry for the misunderstanding Commented Sep 11, 2023 at 17:49
• I'll make it more clear Commented Sep 11, 2023 at 17:50
• All done , my bad ^⁠_⁠^ Commented Sep 11, 2023 at 17:51

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