Sequence Limit

Modern Math 2 Comments

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)

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  1. Pingback: Our Daily Story #8: The Rigorous Mathematician with epsilon-delta | Math Online Tom Circle

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