Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$

Question

Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical $$ \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}} $$ Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that $$ f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} $$ We can see that $$\begin{align} f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\ &= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}} \end{align}$$ And we begin solving by $$\begin{align} f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\ f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\ &= 2^x + f(x + 1) \\ f(x + 1) &= f(x)^2 - 2^x \end{align}$$ At this point I find myself stuck, as I have little experience with recurrence relations.

How would this recurrence relation be solved? Would the method extend easily to $$\begin{align} f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\ f_n(x)^2 &= n^x + f_n(x + 1)~\text ? \end{align}$$

Answer 1

Introduce the notation $[a_0]=\sqrt{a_0}$, and $[a_0,a_1]=\sqrt{a_0+\sqrt{a_1}}$, and so on, including infinite lists: $$[a_0,a_1,a_2,...]=\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}=\sqrt{a_0+[a_1,a_2,\ldots]}.$$ Generally $[a_0,a_1,\ldots]^2 = a_0 + [a_1,a_2,\ldots]$, so for constant-term lists we have a closed-form solution: $$ [x,x,\ldots]^2=x+[x,x,\ldots] \implies [x,x,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{1+4x}. $$ If $b_i \le a_i$ for each $i$, then clearly $[b_0,b_1,\ldots]\le[a_0,a_1,\ldots]$. What happens when a multiplicative factor is introduced? You have $$k[a_0,a_1,a_2,...]=\sqrt{k^2 a_0 + k^2[a_1,a_2,...]}=[k^2a_0,k^4a_1,k^8a_2,\ldots]$$ In your case, $[1,2,4,\ldots]=\sqrt{1+[2,4,8,\ldots]}=\sqrt{1+\sqrt{2}[1,1,1/2,1/16,\ldots]}$. Using the bounds $$ \sqrt{2}=[1,1]\le[1,1,1/2,1/16,\ldots]\le[1,1,1,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{5}, $$ you have $$ 1.732 \approx \sqrt{3} \le [1,2,4,\ldots] \le \sqrt{1+\frac{1+\sqrt{5}}{\sqrt{2}}}\approx 1.813. $$ Tighter bounds can be provided, of course, but this suffices to show that the limit exists.

Answer 2

Let $x_0 = \sqrt{1}$, $x_1 = \sqrt{1 + \sqrt{2}}$, $x_3 = \sqrt{1 + \sqrt{2 + \sqrt{4}}}$, and so on. Then we have:

$$\sqrt{1 + \sqrt{2 + \sqrt{4 + \cdots}}} = \lim_{n \to \infty} x_n$$

Clearly, this sequence is monotonically increasing. It also converges, since we can see each new term appears under $n$ square roots, and hence $| x_n - x_{n - 1} | \propto 2^{-n}$ which should be enough.

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From observation, the value of $x_n - x_{n - 1}$ is a root of a polynomial of order $2^{2n - 3}$. In this sense, a closed-form solution is very unlikely to exist. Not a full answer, but it looks complicated.

Answer 3

Just a comment on such similar cases. From this short answer, $$x+2^n=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\tag{1}$$ Let $N=\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$. Then $$\begin{aligned} N &< \sqrt{1+\sqrt{2+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}} \\ N &< \sqrt{1+\sqrt{5}} \\ &< 1.79890743995\\ \end{aligned}$$ For comparison, $N ≃ 1.78316580926410..$. Also from Eq. $1$, we can get; $$\boxed{1+\sqrt{2}=\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}}$$ $$\begin{aligned} N &< \sqrt{1+\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}} \\ &< \sqrt{2+\sqrt{2}} \\ \end{aligned}$$

Answer 4

This is not an answer to your question, but you may be interested in the following two similar evaluations: $$\sqrt{1+\sqrt{4+\sqrt{16+\sqrt{64+...}}}}=2\tag{1}$$ $$\sqrt{1+2^{-1}\sqrt{1+2^{-2}\sqrt{1+2^{-3}\sqrt{1+...}}}}=\frac{5}{4}\tag{2}$$ To prove $(1)$, one may use the fact that $$2^n+1=\sqrt{4^n+2^{n+1}+1}$$ and $(2)$ can be obtained from $(1)$ by multiplying it by $1/2$ and inverting the first radical.

Answer 5

Since you have recurrence relation $(f(x))^2=2^x+f(x+1)$ you could find an approximate solution by approximating $f(x+1)\approx f(x)$ and then you get quadratic equation for the function $f$ and in doing so you can find an approximate value $f(x)$ for every value of $x$. It is clear from the recurrence relation that the exact expression for $f(x)$ may not be expressible as some combination of the functions that have been studied to this day, but maybe I am wrong.

If you still want to search for closed forms for certain values of $f$ maybe it is better to study expressions of this type that are finite, such as $f(2,x,n) =\sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} +\sqrt {\ldots+\sqrt{2^{x+n}}}}}}}$ and then let $n\to\infty$ to get your $f$.

Answer 6

I'd rather try this: If $$x=\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}},$$ then $$\begin{align}ux&=u\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+u^2\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+\sqrt{2u^4+u^4\sqrt{4+\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+u^8\sqrt{8+\ldots}}}}\\ &=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+\sqrt{8u^{16}+\ldots}}}}\\ \end{align}$$ Thus if $u=\frac1{\sqrt 2}$ then $$ \frac x{\sqrt2}=\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\ldots}}}}$$ The right hand side $y$ has the property that $y^2-\frac12=y$, i.e. can be found by solving a quadratic.