Sequence Limit

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

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

$latex \Updownarrow $

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

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

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

This means:

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

Example:

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

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

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

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

Let's prove it.

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

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

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

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

[Source]: Excellent Introduction in Modern Math:
"A Concise Introduction to Pure Mathematics (3rd Edition)"
by Martin Liebeck
CRC Press @ 2011

Solution 1 (Sequence): Monkeys & Coconuts

Monkeys & Coconuts Problem

Solution 1 : iteration problem => Use sequence
$Latex U_{j} =\frac {4}{5} U_{j- 1} -1 $

(initial coconuts)
$Latex U_0 =k$
Let
$Latex f(x)=\frac{4}{5}(x-1)=\frac{4}{5}(x+4)-4$
$Latex U_1 =f(U_0)=f(k)= \frac{4}{5}(k+4)-4$

$Latex U_2 =f(U_1)=f(\frac{4}{5}(k+4)-4)= \frac{4}{5}((\frac{4}{5}(k+4)-4+4)-4$

$Latex U_2=(\frac{4}{5})^2 (k+4)-4$

$Latex U_3=(\frac{4}{5})^3 (k+4)-4$

$Latex U_4=(\frac{4}{5})^4 (k+4)-4$

$Latex U_5=(\frac{4}{5})^5 (k+4)-4$

Since
$Latex U_5$ is integer  ,
$Latex 5^5 divides (k+4)$
k+4 ≡ 0 mod($Latex 5^5$)
k≡-4 mod($Latex 5^5$)
Minimum {k} = $Latex 5^5 -4$= 3121 [QED]

Note: The solution was given by Paul Richard Halmos (March 3, 1916 – October 2, 2006)