# 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:

A Concise Introduction to Pure Mathematics, Third Edition (Chapman Hall/CRC Mathematics Series)   ## 2 thoughts on “Sequence Limit”

1. tomcircle

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