# Sequence ?

## Hey guys!! sorry cz i didn't use LATEX. Cz i really don't know how to use it. Need help in the first exercise, I have to find all values of a_1 such that the sequence is converge and i have to find the limit too. Thanks alot. https://files.acrobat.com/a/preview/b35761b1-baaf-4004-94b9-b140bef48ec4 the link of the exercise!!!

Feb 5, 2017

$\forall {a}_{1} \in \mathbb{R} {\lim}_{n \to \infty} {a}_{n} = a$

#### Explanation:

Given $a \in \mathbb{R}$, the sequence $\left\{{a}_{n}\right\}$ is defined as:

${a}_{n + 1} = {a}_{n}^{2} + \left(1 - 2 a\right) {a}_{n} + {a}^{2}$

Simplify:

${a}_{n + 1} = {a}_{n}^{2} + {a}_{n} - 2 a \cdot {a}_{n} + {a}^{2} = {a}_{n} + {\left({a}_{n} - a\right)}^{2}$

Passing to the limit:

${\lim}_{n \to \infty} {a}_{n + 1} = {\lim}_{n \to \infty} {a}_{n} + {\lim}_{n \to \infty} {\left({a}_{n} - a\right)}^{2}$

But if the series is convergent:

${\lim}_{n \to \infty} {a}_{n + 1} = {\lim}_{n \to \infty} {a}_{n} = L$

so:

${\lim}_{n \to \infty} {\left({a}_{n} - a\right)}^{2} = 0$

and also:

$\sqrt{{\lim}_{n \to \infty} {\left({a}_{n} - a\right)}^{2}} = 0$

but since $\sqrt{x}$ is a continuous function:

$\sqrt{{\lim}_{n \to \infty} {\left({a}_{n} - a\right)}^{2}} = {\lim}_{n \to \infty} \sqrt{{\left({a}_{n} - a\right)}^{2}} = {\lim}_{n \to \infty} \left\mid {a}_{n} - a \right\mid = 0$

That is:

${\lim}_{n \to \infty} {a}_{n} = a$

independently of how we choose ${a}_{1}$