<s>[INST] Find square roots upto infinte times Evaluate :
$\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$
Is it possible to solve in the following way :
Let
$x=\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$
$x^2= 1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}$  
next ??????  
I don't know
 [/INST] Note that $(x+1)^2=1+x(x+2)$, now taking the principal square root of both sides and re-substituting the resulting expression for $x+1$ multiple times gives us that:
$$(x+1)=\sqrt{1+x(x+2)}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)(x+3)})}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{1+(x+4)....}}}}}$$
$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7}}...}}}}$$
You can use the monotone convergence theorem to prove that this resulting expression converges.
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