Ramanujan's infinite root and its crazy cousins

What people said

Recommends Marcel Besixdouze · 2018-10 · Mathematics Stack Exchange · on “Is $i = e^{\frac{\pi}{2}e^{\frac{\pi}{2}^{.^{.^.}}}}$?”

An infinitary notation like $e^{\frac{\pi}{2}e^{\frac{\pi}{2}e^\frac{\pi\cdots}{2}}}$ requires an interpretation. The usual and implied way is as a limit of a sequence. In this case, the standard sequence interpretation would be as the limit of the sequence: $\left\{e^\frac{\pi}{2},{e^{\frac{\pi}{2}e^\frac{\pi}{2}}},{e^{\frac{\pi}{2}e^{\frac{\pi}{2}e^\frac{\pi}{2}}}},\ldots \right\} \hspace{3cm} (\mbox{I})$ This is a divergent sequence that approaches $+\infty$ (the terms are getting larger and larger, without bound). The interpretation you proposed is: $\left\{e^\frac{\pi i}{2},{e^{\frac{\pi}{2}e^\frac{\pi i}{2}}},{e^{\frac{\pi}{2}e^{\frac{\pi}{2}e^\frac{\pi i}{2}}}},\ldots \right\} \hspace

Cites John Barber · 2016-09 · Mathematics Stack Exchange · on “What is this number: $x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\approx 4.14031...$?”

I was watching this [this video] Mathologer video about Ramanujan's nested radical identity $$ 3 = \sqrt{1 + 2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\, , $$ and I decided to look at this for different sets of "radical coefficients". Using powers of $2$ yields the quantity $$ x = \sqrt{1+2^1\sqrt{1+2^2\sqrt{1+2^3\sqrt{1+...}}}}\, . $$ I can't figure out much about the nature of the number $x$. I can't even prove that it exists, let alone anything analytically about its value. Nevertheless, by looking at the truncation of this expression at different levels in Mathematica, it seems to be the case that: 1) The expression converges relatively quickly. 2) $x \approx 4.14031456214125981180937290...$ I tried

Cites PrincessEev · 2019-02 · Mathematics Stack Exchange · on “Ramanujan's radical and how we define an infinite nested radical”

Introduction: The issue is what "..." really "represents." Typically we use it as a sort of shorthand, as if to say "look, I can't write infinitely many things down, just assume that the obvious pattern holds and goes on infinitely." This idea holds for all sorts of things - nested radicals, infinite sums, continued fractions, infinite sequences, etc. On Infinite Sums: A simple example: the sum of the reciprocals of squares: $$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ...$$ This is a well known summation. It is the Riemann zeta function $\zeta(s)$ at $s=2$ , and is known to evaluate to $\pi^2/6$ (proved by Euler and known as the Basel problem [LINK] ). Another, easier-to-handle summatio

Cites keyoted · 2020-05 · Mathematics Stack Exchange · on “Simplifying Ramanujan's limited infinite root”

Background I was playing around with Ramanujan's infinite root : $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}}$$ You can write any number as this sequence by limiting the number of square roots in this way; $$ r_{n}(1) = \sqrt{f_{n}(x_{i})}\\\ r_{n}(x) = \sqrt{1+ (x_i - x + 2) \times r_{n}(x-1)} $$ Where: $x\rightarrow$ Number of square roots. $n\rightarrow$ The number to write. $f_{n}(x_i)\rightarrow$ A number that makes $r_{n}(x)=n$ a true statement. $x_i\rightarrow$ Initial value for $x$ . For example: $r_{3}(2) = \sqrt{1 + 2 \sqrt{16}} = 3$ , where $f_{3}(2) = 16$ $r_{3}(3) = \sqrt{1+ 2 \sqrt{1 + 3 \sqrt{25}}} = 3$ , where $f_{3}(3) = 25$ $r_{4}(3) = \sqrt{1+ 2 \sqrt{1+ 3 \sqrt{\frac{488