<s>[INST] Sum of Infinite Surds Recently I came across this question. Compute $$\sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}.$$ 
I know how to do it. By equating the above expression to $x$ (i.e., $x = \sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}$) and $x = \sqrt{10+x}$ and so on. But how is this justified? I mean if we take the expression $\sqrt{10+x}$, literally it is equal to $\sqrt{10+\sqrt{10+\sqrt{100 +\dots}}}$ which alters the original expression. Can someone clarify my doubt?
 [/INST] $x=\sqrt{10+\sqrt{10^2+\sqrt{10^4+\cdots}}}$
$\implies \frac x{\sqrt{10}}=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ (This is the well-known Golden Ratio)
so, $\frac x{\sqrt{10}}=\frac{ 1+\sqrt 5}{2}$

Alternatively, $ \frac x{\sqrt{10}}=\sqrt{1+\frac x{\sqrt{10}}}$ 
Squaring we get, $\frac {x^2}{10}=1+\frac x{\sqrt{10}}\implies x^2-\sqrt{10}x-10=0$
$\implies x=\frac{\sqrt 10\pm \sqrt {50}}{2}=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$
or $y=\sqrt{1+y}$ putting $\frac x{\sqrt{10}}=y$
$\implies y^2=1+y\implies y=\frac{1\pm\sqrt5}2\implies x=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$
But $x>0,$ which means $x=\frac{\sqrt{10}(1+\sqrt 5)}{2}$
For the converge, one may check for "Geometric Infinite Surd" here.
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