# Showing a sequence is convergent using the $\epsilon$-$N$ definition

I need to prove that the following sequence converges:

$\lim_{n\rightarrow \infty} \frac{2n^2+3n+1}{n^2+n+1}=2$

So for the proof/solution I have the following:

Let $\epsilon >0$. Then let $N=\frac{1}{\epsilon}$. Then for all $n\geq N$, $|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$

Thus the sequence converges to 2.

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It depends on the class and what's expected of you (even down to your professor and their style really). I've had one real analysis class where that wouldn't be a problem, and another with a professor that required us to prove "every detail" of what we were doing (so making a statement f(n) < g(n) required a side proof using the axioms from the start of the course). In most cases though I'd say it should be fine (with the symbolic adjustment as noted made). – Vilid Sep 13 '12 at 2:45

There is a small error with symbols, it should be like this:

Choose $\epsilon\gt 0$

Let $N=\lceil\frac{1}{\epsilon}\rceil$. Then for all $n\gt N$, $|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$

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Thanks! I noticed that too. – tk2 Sep 13 '12 at 2:43
Then, you are all right :) – Aang Sep 13 '12 at 2:44

The calculations are right, but some wording changes would be useful. We are given an $\epsilon \gt 0$.

We have, for $n \ge 1$, the inequality $$\left|\frac{2n^2+3n+1}{n^2+n+1}-2\right| \lt \frac{1}{n}.$$ Thus if we put $N=\lceil \frac{1}{\epsilon}\rceil$, then $$\left|\frac{2n^2+3n+1}{n^2+n+1}-2\right|\lt \epsilon$$ for every $n\gt N$.

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