<s>[INST] Estimating the nested radical with $1,2,3,4,\dots$ under square root signs 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.
 [/INST] 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 $1$.
$\implies 1 < \sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} <  \sqrt{ 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$
$\implies 1 < \sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < \dfrac{1 + \sqrt{5}}{2}$
$\implies 2 < 2\sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < 1 + \sqrt{5}$
$\implies 2 < \sqrt{4 + \sqrt{5+\sqrt{6+\sqrt{7+\cdots}}}} < 1 + \sqrt{5}$
$\implies 5 < 3 + \sqrt{4 + \sqrt{5+\sqrt{6+\sqrt{7+\cdots}}}} < 4 + \sqrt{5}$
$\implies \sqrt{1 + \sqrt{2 + \sqrt{5}}} < \sqrt{ 1 + \sqrt{2 + \sqrt{3 + \sqrt{4 + \cdots}}}} < \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{5}}}}$
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