# Using the definition of convergence, how do you prove that the sequence lim (n + 2)/ (n^2 - 3) = 0 converges from n=1 to infinity?

Dec 22, 2017

You apply the deffinition and then simplify

#### Explanation:

First of all you need to proof that $\forall \epsilon > 0 \exists k \in \mathbb{N} : \forall n \ge k , | {x}_{n} - 0 | < \epsilon$

So:

$| {x}_{n} - 0 | : = | \frac{n + 2}{{n}^{2} - 3} - 0 | = | \frac{n + 2}{{n}^{2} - 3} |$ here you can say that the sequence converges to 0 from $n = 1 \iff$ it converges to 0 from $n = 3$.
So now you have $n \ge 3$ and $| {x}_{n} - 0 | = | \frac{n + 2}{{n}^{2} - 3} | \le | \frac{n + 2}{{n}^{2} - 4} | = | \frac{n + 2}{\left(n + 2\right) \left(n - 2\right)} | = | \frac{1}{n - 2} | = \frac{1}{n - 2} < \epsilon$ if you choose any $k$ that verifies $\frac{1}{k - 2} < \epsilon$ so $k > \frac{1}{\epsilon} + 2$