<s>[INST] Proving $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$ by induction How can I prove by induction that $\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}} < 3$ for $n\geq 1$? My guess is that there must be another form to express the sum of nested square roots, but I don't know how to find it.
 [/INST] $$(1) \quad x=k+\sqrt x$$
$$x=\sqrt {k+\sqrt x}$$
continuing the recursion...
$$x=\sqrt {k+\sqrt {k+\sqrt {k+...}}}$$
Thus (1) is the equivalent expression.
Solve for x with $k=1$
$$x=\phi$$
Thus, x equals the golden ratio. Multiply the expression by c...
$$c \cdot x=c \cdot \sqrt {k+\sqrt {k+\sqrt {k+...}}}$$ 
$$c \cdot x=\sqrt {c^2 \cdot k+\sqrt {c^4 \cdot k+\sqrt {c^8 \cdot k+...}}}$$ 
Choose c to equal $2^{1/4}$ (This is where you use induction to prove this works)
Prove $$2^{2^n/4} \ge n$$
For $n=1 \ $, $ \ \sqrt 2 \gt 1$ so we have a base case
Inductive step, assume the above holds for $n=k$
Let, $n=k+1$
$$2^{2^k/4} \ge k$$
$$2 \cdot 2^{2^k/4} \ge 2 \cdot k$$
$$ 2^{2 \cdot 2^{k}/4} \ge 2 \cdot k$$
$$2^{2^{k+1}/4} \ge k+1$$
assuming k is greater than 1, thus the above holds for all n.
$$2^{1/4} \cdot \phi=\sqrt {\sqrt 2+\sqrt {2+\sqrt { 4+\sqrt {16+...}}}}$$ 
Thus the inequality for the radical is given by
$$\phi \le S \le 2^{1/4} \cdot \phi$$
$$1.618... \le S \le 1.924...$$
I personally guess the radical to be approximately equal to $1.771$...
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