How to calculate $\lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$.

Question

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?

@JessePFrancis I want $t^{2} = 2t^{4} = 4t^{8} = 8t^{16} = ...$. — T. Golden, Nov 16 '15 at 11:52
My calculator's telling me $s=1.78317$, which doesn't match what you have. — Akiva Weinberger, Nov 16 '15 at 11:52
@AkivaWeinberger Yes, that is why I am here to ask. I do not know why this is, though. — T. Golden, Nov 16 '15 at 11:52
Maybe it's because, when you simplify $st$, you do infinitely many operations at one. I agree that $st=\sqrt{t^2+t^2\sqrt{\dotsb}}$, and that $st=\sqrt{t^2+\sqrt{2t^4+t^4\sqrt{\dotsb}}}$, etc., but I'm not sure that you can assume that it's equal to what you posted above, since you're doing infinitely many operations. — Akiva Weinberger, Nov 16 '15 at 11:57
The correct way to put it is: Find the limit for $n\to +\infty$ of $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ...+\sqrt{2^n}}}}}$ — Jimmy R., Nov 16 '15 at 11:59
If I'm reading this question correctly, it doesn't converge so the formal manipulations are meaningless.... http://math.stackexchange.com/questions/322690/convergence-of-nested-radicals — Myself, Nov 16 '15 at 12:04
More blatant error: 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. — Akiva Weinberger, Nov 16 '15 at 12:04
The coefficients double each time, but you seem to have thought that they square each time. I'm guessing that this is the main error, and that the infinitely many operations bit is probably fine. — Akiva Weinberger, Nov 16 '15 at 12:06
@Myself, Excel says terms beyond $x=8$ get very very close to $1.783166$ (I tried up to $x=20$), it should converge. — Jesse P Francis, Nov 16 '15 at 12:27

score 1 · Accepted Answer · answered Nov 16 '15 at 12:11

1

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.

answered Nov 16 '15 at 12:11

Akiva Weinberger

23,062

Thank you for your answer. Would you be able to help me obtain a value of $t$ that would work, if one exists. – T. Golden Nov 16 '15 at 12:13
I don't think your plan would work, for any value of $t$. – Akiva Weinberger Nov 16 '15 at 12:24
I think that, if you do powers of $4$ instead of powers of $2$, the answer is equal to $2$. That is,$$\sqrt{1+\sqrt{4+\sqrt{16+\sqrt{64+\dotsb}}}}=2$$ – Akiva Weinberger Nov 16 '15 at 12:28
Ugh! Never mind -_- – Myself Nov 16 '15 at 13:15
@T.Golden Relevant forum – Akiva Weinberger Nov 16 '15 at 13:48

How to calculate $\lim_{x \to \infty}{\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + ... \sqrt{2^{x}}}}}}}$.

1 Answers1