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I require some assistance in proving the following inequality:

$$\sqrt{1 + \sqrt{2 + \sqrt{5}}} < \sqrt{ 1 + \sqrt{2 + \sqrt{3 + \sqrt{4 + \cdots}}}} < \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{5}}}}$$

I'm not quite sure how to create a rigorous enough argument here, help would be appreciated.

TI82
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1 Answers1

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I was just curious if this was a way it could be resolved:

$0 < \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}} < \sqrt{ 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$

This is true because each term in the middle radical is less than $1$.

$\implies 1 < \sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < \sqrt{ 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$

$\implies 1 < \sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < \dfrac{1 + \sqrt{5}}{2}$

$\implies 2 < 2\sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < 1 + \sqrt{5}$

$\implies 2 < \sqrt{4 + \sqrt{5+\sqrt{6+\sqrt{7+\cdots}}}} < 1 + \sqrt{5}$

$\implies 5 < 3 + \sqrt{4 + \sqrt{5+\sqrt{6+\sqrt{7+\cdots}}}} < 4 + \sqrt{5}$

$\implies \sqrt{1 + \sqrt{2 + \sqrt{5}}} < \sqrt{ 1 + \sqrt{2 + \sqrt{3 + \sqrt{4 + \cdots}}}} < \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{5}}}}$

Maazul
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  • You have found a consequence of the statement. The question was to prove the original statement. – Fly by Night Jun 08 '13 at 19:26
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    Edited for completeness. – Maazul Jun 08 '13 at 23:02
  • If the last inequality is correct then you need to reverse the direction of all of you implications. You started with what you wanted to end up with. – Fly by Night Jun 09 '13 at 11:07
  • Hello, I have a similar question about this when proving inequalities. If we go onto find a consequence of the original statement, can the original statement be validated if using an if and only if argument? – user78416 Jun 09 '13 at 11:20