Question

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.

Answer 1

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+sqrty)` so

when `n->oo` then `a_n->y` and

`y=sqrt(c+sqrt(y))` or

`y^2=c+sqrty->(y^2-c)^2=y` or

calling `z=sqrty` 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))->a_(n+1)^2=c+sqrt(a_n)` then if `c > 0`

`a_(n+1)^2>sqrt(a_n)` Applying `log` to both sides

`2log(a_(n+1)) > 1/2log(a_n)` or

`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))`

(remember that `a_1= sqrt(3) > 1`) so

`a_(n+1)^2-a_(n+1)-c < 0` or

`a_(n+1) < 1/2 (1 + sqrt[1 + 4 c])`