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This question is from Arthur Mattuck's "Introduction to Analysis", chapter 5, problem 5-7.
Define a sequence recursively by [tex]a_{n+1}=\sqrt{2a_{n}}[/tex], [tex]a_{0}>0[/tex].
(a) Prove that for any choice of [tex]a_{0}>0[/tex], the sequence is monotone and bounded.
None
I've rewritten the sequence as
[tex]a_n = \sqrt{2{\sqrt{2\sqrt{2\sqrt{2...\sqrt{2a_0}}}}}[/tex],
and taken the ratio of [tex]a_{n+1}[/tex] and [tex]a_{n}[/tex], which leads me to the expression
[tex]\frac{a_{n+1}}{a_n} = 2^{(1/2)^{n+1}}\sqrt{a_0}[/tex].
From computing the sequence for a few initial values of [tex]a_0[/tex], I've been able to determine that the sequence is decreasing if [tex]a_0>2[/tex] and increasing if [tex]0<a_0<2[/tex]. However, I'm not sure how to show this from this ratio, i.e. how does
[tex]2^{(1/2)^{n+1}}\sqrt{a_0}<1[/tex]
which indicates that the sequence is decreasing, turn into [tex]a_0>2[/tex]?
The rest of the problem comes easy enough. I'm able to show that it is bounded and that the limit is 2 (in part b of the question).
Homework Statement
Define a sequence recursively by [tex]a_{n+1}=\sqrt{2a_{n}}[/tex], [tex]a_{0}>0[/tex].
(a) Prove that for any choice of [tex]a_{0}>0[/tex], the sequence is monotone and bounded.
Homework Equations
None
The Attempt at a Solution
I've rewritten the sequence as
[tex]a_n = \sqrt{2{\sqrt{2\sqrt{2\sqrt{2...\sqrt{2a_0}}}}}[/tex],
and taken the ratio of [tex]a_{n+1}[/tex] and [tex]a_{n}[/tex], which leads me to the expression
[tex]\frac{a_{n+1}}{a_n} = 2^{(1/2)^{n+1}}\sqrt{a_0}[/tex].
From computing the sequence for a few initial values of [tex]a_0[/tex], I've been able to determine that the sequence is decreasing if [tex]a_0>2[/tex] and increasing if [tex]0<a_0<2[/tex]. However, I'm not sure how to show this from this ratio, i.e. how does
[tex]2^{(1/2)^{n+1}}\sqrt{a_0}<1[/tex]
which indicates that the sequence is decreasing, turn into [tex]a_0>2[/tex]?
The rest of the problem comes easy enough. I'm able to show that it is bounded and that the limit is 2 (in part b of the question).