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.

1704 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-05T21:25:08

See below.

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.

enter image source here

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])#