Recently I was studying functional equation, in which I came across this paragraph about Ramanujan's Nested Radical in Christopher G. Small's book [Functional Equations and How to Solve Them][1].
$$ \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3. $$ assuming $$ f(x) = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}=3.$$ $$ [f(x)]^2 = 1 + x f(x+1) $$ by degree analysis we get $ f(x) = x+1 $ (by assuming $f(x) = ax+b$ ,thus by substituting a,b =1)
Next comes the main part i.e the author derives an inequality before proceeding to proof.
Now I'm providing the derivation of inequality as it is
for all $x \geq 1$ $$ f(x) \geq {\sqrt{x\sqrt{{x\sqrt{....}}}}} $$ $$ = x^{(1/2)+(1/4)+(1/8)+.....} $$ $$ = x $$ $$ \geq (x+1)/2 $$
Bounding on other side, We have
$$ f(x) \leq {\sqrt{(x+1)\sqrt{{(x+2)\sqrt{{ (x+3)}....}}}}} $$ $$ f(x) \leq {\sqrt{(x+1)\sqrt{{2(x+1)\sqrt{{ 3(x+1)}....}}}}} $$ $$ f(x) = (x+1) {\sqrt{1 \sqrt{{2\sqrt{{ 3}....}}}}} $$ $$ f(x) \leq (x+1) {\sqrt{1 \sqrt{{2\sqrt{{ 4}....}}}}} $$ $$ f(x) = 2(x+1) $$
Thus
$$ (x+1)/2 \leq f(x) \leq 2(x+1) $$
I can't understand the proceeding of Inequality. Is there any inequality used? Is it an approximation then how do it proceeds as such? (This inequality is used further to prove,(regards Thesimplifire for pointing out a typo) and available on the next page of book)
Small, Christopher G., Functional equations and How to solve them, Problem Books in Mathematics. New York, NY: Springer (ISBN 978-0-387-34539-0/pbk). xii, 129 p. (2007). ZBL1152.39300.