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