Evaluate an infinite staircase of square roots of square roots plus square roots .

Question

This is an expression involving a concatenation of overlapping square roots of square roots,

 

x =  √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

              + √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

                        + √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

                                  + √16¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

                                            + etc., whereby all radicals encompass ALL subsequent radicands (which are themselves radicals) for any number of iterations N. I want to be evaluate the limit as N approaches ∞.

Answer 1

There are several way to find that limit. One of them is as following:

x_n=√16+√16+..=√16+x_n-1

x_n=√(16+x_n-1)  (LIM)

Now, if there exists (!) Lim_x→∞x_n =a,  Then Lim_x→∞x_n-1 =a as  well

Now taking limit from both sides of (LIM) gives you

a=√(16+a)

a²=16+a

a²-a-16=0

If you solve this quadratic equation, you get 

a₁=(1/2) (1 -√65), a₂=(1/2) (1 + √65).

So,

Lim_x→∞x_n =(1/2) (1 + √65).  (I guess, you know why not not a₁).

 

Note: When finding the limit this way, somehow you have to show that it exists and and bounded. In principal, you have to clarify that in the  first place.