Evaluating nested radicals

We multiply $\displaystyle S$ by $\displaystyle \frac{1}{\sqrt2}$ and try to push it inside the nested radicals:

$\displaystyle \frac{S}{\sqrt2} = \frac{1}{\sqrt2} \times \sqrt{\frac{2}{2^{2^{0}}} + \sqrt{\frac{2}{2^{2^{1}}} + \ldots}}$

$\displaystyle \Rightarrow \frac{S}{\sqrt2} = \sqrt{\frac{1}{2} \times \bigg( \frac{2}{2^{2^{0}}} + \sqrt{\frac{2}{2^{2^{1}}} + \ldots} \bigg)}$

$\displaystyle \Rightarrow \frac{S}{\sqrt2} = \sqrt{ \frac{2}{2^{2^{1}}} + \frac{1}{2} \times \bigg( \sqrt{\frac{2}{2^{2^{1}}} + \ldots} \bigg) }$

$\displaystyle \Rightarrow \frac{S}{\sqrt2}= \sqrt{ \frac{2}{2^{2^{1}}} + \sqrt{ \frac{2}{2^{2^{2}}} + \frac{1}{2^{2}} \times (\ldots) } }$

We realise that this sequence is the same as $\displaystyle S^2$ without it's first term, so:

$\displaystyle \Rightarrow \frac{S}{\sqrt2} = S^2 - \frac{2}{2^{2^{0}}}$

$\displaystyle \Rightarrow \sqrt{2} S^2 - S - \sqrt{2} = 0$

Solving the quadratic, we get $S = \sqrt2, - \frac{1}{\sqrt{2}}$ .

But, since $\displaystyle S$ is positive, $\displaystyle \big(S = -\frac{1}{\sqrt2}\big)$ is discarded. Hence:

$\displaystyle \Rightarrow S = \sqrt{A} = \sqrt2$

$\displaystyle \Rightarrow A = \boxed{2}$

Let $y$ be the nested radical, then we have:

$\begin{aligned} y & = \sqrt{\frac{2}{2^{2^0}}+\sqrt{\frac{2}{2^{2^1}}+\sqrt{\frac{2}{2^{2^2}}+\sqrt{\frac{2}{2^{2^3}}+...}}}} \\ & = \sqrt{\frac{2}{2}+\sqrt{\frac{2}{2^2}+\sqrt{\frac{2}{2^4}+\sqrt{\frac{2}{2^8}+...}}}} \\ & = \frac{1}{\sqrt{2}}\color{#3D99F6}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} \quad \quad \small \text{Let } x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}} \\ \Rightarrow y & = \frac{\color{#3D99F6}x}{\sqrt{2}} \\ \Rightarrow x & = \sqrt{2+x} \\ x^2 & = 2 + x \\ x^2 - x - 2 & = 0 \\ (x-2)(x+1) & = 0 \\ \Rightarrow x & = 2 \quad \quad \small \color{#3D99F6}{x > 0} \\ \Rightarrow y & = \frac{x}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \\ & \\ \Rightarrow A & = \boxed{2} \end{aligned}$

We can solve this by approximation method also. If we see the first term 2/2^2^0=1 and the remaining all term cannot be greater than 1 . ao, it means the total value cannot be greater than 2. So, The answer is 2.