# Converge 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/a06b02a6-85f9-468c-97ca-9d3ff166c14b THE LINK.

Feb 5, 2017

See below.

#### Explanation:

1) Giving

${a}_{n + 1} = \sqrt{c + \sqrt{{a}_{n}}}$ and making substitutions...

$y = \sqrt{c + \sqrt{c + \sqrt{c + \cdots}}} = \sqrt{c + \sqrt{y}}$ so

when $n \to \infty$ then ${a}_{n} \to y$ and

$y = \sqrt{c + \sqrt{y}}$ or

${y}^{2} = c + \sqrt{y} \to {\left({y}^{2} - c\right)}^{2} = y$ or

calling $z = \sqrt{y}$ we have

${z}^{4} = c + z$. This polynomial has two real roots and two conjugate complex roots.

One of the real roots is the answer for the limit value.

Considering $c = 14$ we obtain $z = 2$ and $y = 4$

The attached plot shows the sequence elements converging to the final value.

Now the convergence

${a}_{n + 1} = \sqrt{c + \sqrt{{a}_{n}}} \to {a}_{n + 1}^{2} = c + \sqrt{{a}_{n}}$ then if $c > 0$

${a}_{n + 1}^{2} > \sqrt{{a}_{n}}$ Applying $\log$ to both sides

$2 \log \left({a}_{n + 1}\right) > \frac{1}{2} \log \left({a}_{n}\right)$ or

$\log \left({a}_{n + 1} / \left({a}_{n}\right)\right) > 0 \to {a}_{n + 1} / \left({a}_{n}\right) > 1 \to {a}_{n + 1} > {a}_{n}$

and also in consequence

${a}_{n + 1} < \sqrt{c + \sqrt{{a}_{n + 1}}} < \sqrt{c + {a}_{n + 1}}$

(remember that ${a}_{1} = \sqrt{3} > 1$) so

${a}_{n + 1}^{2} - {a}_{n + 1} - c < 0$ or

${a}_{n + 1} < \frac{1}{2} \left(1 + \sqrt{1 + 4 c}\right)$