# convergent sequence

• Oct 9th 2007, 09:31 AM
Dili
convergent sequence
a= (n^2-5n+6)/(n^2+n+1)

using the definition only prove the sequence(a) is convergent and find its limit?

what does this prove by definition means? is it same as when proving a limit?

any help is welcome. thanks
• Oct 9th 2007, 09:36 AM
Jhevon
Quote:

Originally Posted by Dili
a= (n^2-5n+6)/(n^2+n+1)

using the definition only prove the sequence(a) is convergent and find its limit?

what does this prove by definition means? is it same as when proving a limit?

any help is welcome. thanks

Definition: Let $\displaystyle \{ a_n \}$ be a sequence of real numbers. We say "$\displaystyle \{ a_n \}$ converges to $\displaystyle L$" if for every $\displaystyle \epsilon > 0$, there exists an $\displaystyle N \in \mathbb {N}$, such that $\displaystyle n > N$ implies $\displaystyle |a_n - L |< \epsilon$. We write $\displaystyle \lim_{n \to \infty}a_n = L$ when this happens.
• Oct 9th 2007, 03:34 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
Definition: Let $\displaystyle \{ a_n \}$ be a sequence of real numbers. We say "$\displaystyle \{ a_n \}$ converges to $\displaystyle L$" if for every $\displaystyle \epsilon > 0$, there exists an $\displaystyle N \in \mathbb {N}$, such that $\displaystyle n > N$ implies $\displaystyle |a_n - L |< \epsilon$. We write $\displaystyle \lim_{n \to \infty}a_n = L$ when this happens.

It seems the poster is asking how to use this definition to prove his result. Not the actual definition.

Quote:

a= (n^2-5n+6)/(n^2+n+1)
Experimentation suggests that $\displaystyle \lim a_n = 1$. We want to show $\displaystyle |a_n - 1|$ can be made as small as we want. Meaning,
$\displaystyle \left| \frac{n^2 - 5n+6}{n^2+n+1} - 1 \right| = \left| \frac{-6n+5} {n^2+n+1} \right| \leq \frac{6n+5}{n^2+n+1} \leq \frac{6n+5n}{n^2} = \frac{11}{n}$

What I wrote is a hint, complete the arugment thyself.