# 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.

• 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). Commented Sep 13, 2012 at 2:45

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 $n\gt N$.

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 $n\gt N$, $|\frac{2n^2+3n+1}{n^2+n+1}-2| = |\frac{n-1}{n^2+n+1}| < \frac{1}{n} < \frac{1}{N} = \epsilon$

• Then, you are all right :)
– Aang
Commented Sep 13, 2012 at 2:44