Mathematics Stack Exchange
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Questions tagged [nested-radicals]

In algebra, a nested radical is a radical expression (one containing a square root sign, cube root sign, etc.) that contains (nests) another radical expression.

In algebra, a nested radical is a radical expression (one containing a square root sign, cube root sign, etc.) that contains (nests) another radical expression. Reference: Wikipedia

Some nested radicals can be rewritten in a form that is not nested. Rewriting a nested radical in this way is called denesting.

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All real numbers in $[0,2]$ can be represented as $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$

I would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$. I'll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted by brackets. $$2=\sqrt{2 +…
Yuriy S
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Is $\sqrt {2 \sqrt {3 \sqrt {4 \ldots}}}$ algebraic or transcendental?

I thought it was easy to show that $\sqrt {2 \sqrt {3 \sqrt {4 \ldots}}}$ is irrational, but found a gap in my proof. Simple finite approximations show the denominator cannot be small, though, strongly suggesting irrationality. However, can it be…
user2566092
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$\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$

(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12) For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x > 0$. Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been…
cnuulhu
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Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$

The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for…
Paramanand Singh
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How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$

Question: Show that $$A=\lim_{n\to \infty}\sqrt{1+\sqrt{\dfrac{1}{2}+\sqrt{\dfrac{1}{3}+\cdots+\sqrt{\dfrac{1}{n}}}}}$$ exists, and find the best estimate limit $A$. It is easy to show…
math110
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$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation

Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator. $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
user2378
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Strategies to denest nested radicals $\sqrt{a+b\sqrt{c}}$

I have recently read some passage about nested radicals, I'm deeply impressed by them. Simple nested radicals $\sqrt{2+\sqrt{2}}$,$\sqrt{3-2\sqrt{2}}$ which the later can be denested into $1-\sqrt{2}$. This may be able to see through easily, but how…
JSCB
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How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists?

I would like to investigate the convergence of $$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt\ldots}}}}$$ Or more precisely, let $$\begin{align} a_1 & = \sqrt 1\\ a_2 & = \sqrt{1+\sqrt2}\\ a_3 & = \sqrt{1+\sqrt{2+\sqrt 3}}\\ a_4 & =…
MJD
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Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $.

How does one prove the following limit? $$ \lim_{n \to \infty} \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}} = 3. $$
anonymous
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Find the value of $\sqrt{10\sqrt{10\sqrt{10...}}}$

I found a question that asked to find the limiting value of $$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$If you make the substitution $x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$ it simplifies to $x=10\sqrt{x}$ which has solutions…
Anthony Muleta
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How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$

Ramanujan stated this radical in his lost notebook: $$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$$ I don't have any idea on how to prove this. Any help appreciated. Thanks.
Shobhit
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Irrationality of $\sqrt{2\sqrt{3\sqrt{4\cdots}}}$

In this question it is stated that Somos' quadratic recurrence constant $$\alpha=\sqrt{2\sqrt{3\sqrt{4\sqrt{\cdots}}}}$$ is an irrational number. [update: the author of that question is no longer claiming to have a proof of this] This fact seems by…
Mizar
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Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$

Find the limit following: $$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$ P.S I tried to find the value of $\:L$, but I found myself stuck into the abyss of…
Iloveyou
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Why is this series of square root of twos equal $\pi$?

Wikipedia claims this but only cites an offline proof: $$\lim_{n\to\infty} 2^n \sqrt{2-\sqrt{2+\cdots+ \sqrt 2}} = \pi$$ for $n$ square roots and one minus sign. The formula is not the "usual" one, like Taylor series or something like that, so I…
chx
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Converting sums of square-roots to nested square-roots

When solving different equations, I have realised, that some roots containing only arithmetic operations and square roots (4th, 8th roots too, because they can be represented using only square roots) can be converted to nested square roots form.…
Somnium
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