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How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ & $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?

I understand that $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$


$$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$

How to Proceed further?

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If compute has the sense of determine the value of, I see no reason why this should be possible (do you have any?). On the other hand, one can show that these iterative definitions indeed converge (is this in fact your question?). – Did Sep 23 '12 at 15:41
I understand that the above expressions converge.But I want to know how to determine the value of above expressions. – Ranjan Yajurvedi Sep 23 '12 at 15:59
Then we are back to my first question: do you have any reason to suspect that one can determine their value? – Did Sep 23 '12 at 16:01
Yes. Because both the above expressions converge(to my knowledge). If we take the first expression though the magnitude of the terms increase with deeper nesting so does the power which keeps on decreasing i.e., 4th root, 8th root, etcetera – Ranjan Yajurvedi Sep 23 '12 at 16:08
The same applies to $\sum\limits_{n\geqslant1}10^{-n!}$, which has no known expression. – Did Sep 23 '12 at 16:14

3 Answers 3


Where $C \approx 1.7579$ (OEIS A072449) is the nested radical constant. No closed form of this constant is known.


where $\sigma \approx 1.6616$ (OEIS A112302) is Somos's Quadratic Recurrence Constant.

$\sigma$ has an alternate form (I hesitate to call it a "closed form") of $$\sigma = \exp\left[-2^n \frac{\partial \operatorname{Li_n\left(\frac{1}{2}\right)}}{\partial n}\bigg|_{n=0}+\frac{1}{2} \frac{\partial \Phi\left(\frac{1}{2},s,n+1\right)}{\partial s}\bigg|_{s=0}\right]$$

where $\operatorname{Li}_n(z) = \sum^\infty_{n=1} \frac{z^k}{k^n}$ is the polylogarithm and $\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(a+k)^s}$ is the Lerch transcendent, and

$$\sigma = \exp\left[\int_0^1 \frac{1-x}{(x-2)\log x}\, \mathrm{d}x\right]$$

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For the first:

Well, if it converges, we may compute its logarithm. Thus, $\log x = \sum_{k=1}^\infty \frac{\log k}{2^k}$ Now, looking at this, it is evident that this actually DO converge, (the numerators are eventually dominated by $1.5^k$ so $\log x < K + \sum_{k=1}^\infty \frac{1.5^k}{2^k}$ for some finite constant $K.$ Now, this sum is geometric and converges).

Now, computing the VALUE of this is trickier, and it is most probably not a rational number or some "nice" expression. In fact, $x$ converges to about 1.66169 (Mathematica), and is the exponential of some expression involving a Polylogarithm.

EDIT: Mixing recursion and different arithmetic operations usually result in constants that have no nice expressions. For example, $1/F_1 + 1/F_2 + 1/F_4+\dots$ converges to a number which is unknown to be rational or not, irrational $F_i$ is the i'th Fibonacci number,

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Nice example in the Edit! Do you know any reference for it? – Did Sep 23 '12 at 17:03
@did This may be what you want. However, it seems the closed form $$\frac{7-\sqrt{5}}{2}$$ is in fact irrational! – Argon Sep 23 '12 at 17:24
I added link to what I meant, seems like I remembered it wrong, but still, there is no known nice closed form of the sum of reciprocal fibonacci numbers. – Per Alexandersson Sep 23 '12 at 20:32
@Paxinum That may be. The sum of the reciprocals of the Fibonacci numbers converges to a constant known as the reciprocal Fibonacci constant ($\psi \approx 3.35988$), which has been shown to be irrational. – Argon Sep 23 '12 at 20:37

Let $A=(1^{1/2})(2^{1/4})(3^{1/8})...\implies \log A= \sum_{i=1}^{\infty}\frac{\log i}{2^i}$ and this sum converges to $\approx 0.507834$

$\implies (1^{1/2})(2^{1/4})(3^{1/8})...$ converges to $\approx e^{0.507834}$.

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The sum converges! – Per Alexandersson Sep 23 '12 at 16:15
@Paxinum,@did: i rectified the answer – Aang Sep 23 '12 at 17:00

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