Albert G.

asked • 01/29/18# Evaluate an infinite staircase of square roots of square roots plus square roots .

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 ∞.

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## 1 Expert Answer

Bobosharif S. answered • 01/29/18

Tutor

4.4
(32)
Mathematics/Statistics Tutor

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

a=√(16+a)

a

^{2}=16+aa

^{2}-a-16=0If you solve this quadratic equation, you get

a

_{1}=(1/2) (1 -√65), a_{2}=(1/2) (1 + √65).So,

Lim

_{x→∞}x_{n}=(1/2) (1 + √65). (I guess, you know why not not a_{1}).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.

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Albert G.

01/30/18