Converge sequence !!!?

Question

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.

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Answer 1 · 2017-02-05T21:25:08

See below.

1) Giving

$a_{n + 1} = \sqrt{c + \sqrt{a_{n}}}$ $a_(n+1)=sqrt(c+sqrt(a_n))$ and making substitutions...

$y = \sqrt{c + \sqrt{c + \sqrt{c + \dots}}} = \sqrt{c + \sqrt{y}}$ $y = sqrt(c+sqrt(c+sqrt(c+cdots))) = sqrt(c+sqrty)$ so

when $n \to \infty$ $n->oo$ then $a_{n} \to y$ $a_n->y$ and

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

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

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

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

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

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

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

enter image source here

Now the convergence

$a_{n + 1} = \sqrt{c + \sqrt{a_{n}}} \to a_{n + 1}^{2} = c + \sqrt{a_{n}}$ $a_(n+1)=sqrt(c+sqrt(a_n))->a_(n+1)^2=c+sqrt(a_n)$ then if $c > 0$ $c > 0$

$a_{n + 1}^{2} > \sqrt{a_{n}}$ $a_(n+1)^2>sqrt(a_n)$ Applying $log$ $log$ to both sides

$2 log (a_{n + 1}) > \frac{1}{2} log (a_{n})$ $2log(a_(n+1)) > 1/2log(a_n)$ or

$log (\frac{a_{n + 1}}{a_{n}}) > 0 \to \frac{a_{n + 1}}{a_{n}} > 1 \to a_{n + 1} > a_{n}$ $log(a_(n+1)/(a_n)) > 0->a_(n+1)/(a_n) > 1->a_(n+1) > a_n$

and also in consequence

$a_{n + 1} < \sqrt{c + \sqrt{a_{n + 1}}} < \sqrt{c + a_{n + 1}}$ $a_(n+1) < sqrt(c+sqrt(a_(n+1))) < sqrt(c+a_(n+1))$

(remember that $a_{1} = \sqrt{3} > 1$ $a_1= sqrt(3) > 1$ ) so

$a_{n + 1}^{2} - a_{n + 1} - c < 0$ $a_(n+1)^2-a_(n+1)-c < 0$ or

$a_{n + 1} < \frac{1}{2} (1 + \sqrt{1 + 4 c})$ $a_(n+1) < 1/2 (1 + sqrt[1 + 4 c])$