Sequence Limit

Definition: $\text{Sequence } (a_n)$
has limit a

$\boxed{\forall \varepsilon >0, \exists N, \forall n \geq N \text { such that } |(a_n) -a| < \varepsilon}$

$\Updownarrow$

$\displaystyle \boxed{ \lim_{n\to\infty} (a_n) = a }$

What if we reverse the order of the definition like this:

∃ N such that ∀ε > 0, ∀n ≥ N,
$|(a_n) -a| < \varepsilon$

This means:

$\boxed {\forall n \geq N, (a_n) = a }$

Example:

$\displaystyle (a_n) = \frac{3n^{2} + 2n +1}{n^{2}-n-3}$

$\displaystyle\text{Prove: } (a_n) \text { convergent? If so, what is the limit ?}$

Proof:
$\displaystyle (a_n) = 3 + \frac{5n +10}{n^{2}-n-3}$

$n \to \infty, (a_n) \to 3$

Let’s prove it.

$\text {Let } \varepsilon >0$
$\text{Choose N such that } \forall n \geq N,$
$\displaystyle |(a_n) -3| = \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \varepsilon$

$\text{Simplify: } \displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr|$
$\text{Let } n > 10$
$\displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \frac{6n}{\frac{1}{2}n^{2}}= \frac{12}{n} < \varepsilon$

$\text{Choose } N = \max (10, \frac{12}{\varepsilon})$
$\displaystyle\forall n \geq N, |(a_n) -3 | < \frac{12}{n} < \varepsilon$

Therefore,
$\displaystyle \boxed{ \lim_{n\to\infty} (a_n) = 3 }$ [QED]

[Source]: Excellent Introduction in Modern Math: