<s>[INST] How to calculate $\lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$. Let $s = \lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$.
$$st = t\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ...}}}} = \sqrt{t^{2} + \sqrt{2t^{4} + \sqrt{4t^{8} + \sqrt{8t^{16} + ...}}}}$$
Let $2t^{4} = t^{2}$:
$$2t^{2} = 1$$
$$t^{2} = \frac{1}{2}$$
$$t = \frac{1}{\sqrt{2}}$$
$$\frac{s}{\sqrt{2}} = \sqrt{\frac{1}{2} + \sqrt{\frac{1}{2} + \sqrt{\frac{1}{2} + \sqrt{\frac{1}{2} + ...}}}}$$
It can be shown that:
$$\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + ...}}}} = \frac{1 \pm \sqrt{1 + 4x}}{2}$$
Therefore:
$$\frac{s}{\sqrt{2}} = \frac{1 \pm \sqrt{3}}{2}$$
$$s = \frac{\sqrt{2} \pm \sqrt{6}}{2}$$
Is this correct?
 [/INST] If $t=\frac1{\sqrt2}$, then $t^2=\frac12$, and $2t^4=\frac12$, but $4t^8=\frac14$. Your expression for $\frac s{\sqrt2}$ isn't composed entirely of $\frac12$s.
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