Linked Questions

41
votes
5answers
3k views

Ramanujan's radical and how we define an infinite nested radical [duplicate]

I know it is true that we have $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$ The argument is to break the nested radical into something like $$3 = \sqrt{9}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{...
45
votes
4answers
17k views

Evaluating the nested radical $ \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} $.

How does one prove the following limit? $$ \lim_{n \to \infty} \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}} = 3. $$
7
votes
2answers
23k views

Show that $\sqrt{2+\sqrt{2+\sqrt{2…}}}$ converges to 2 [duplicate]

Consider the sequence defined by $a_1 = \sqrt{2}$, $a_2 = \sqrt{2 + \sqrt{2}}$, so that in general, $a_n = \sqrt{2 + a_{n - 1}}$ for $n > 1$. I know 2 is an upper bound of this sequence (I proved ...
22
votes
1answer
2k views

Nested Square Roots $5^0+\sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt\dots}}}$

How would one go about computing the value of $X$, where $X=5^0+ \sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt{5^8+\sqrt{5^{16}+\sqrt{5^{32}+\dots}}}}}}$ I have tried the standard way of squaring then trying ...
5
votes
6answers
1k views

how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$ [duplicate]

Possible Duplicate: Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$ we know, if $x=\...
8
votes
3answers
2k views

Evaluating $\sqrt{6+\sqrt{6+\cdots}}$

Tough as introduction to analysis for beginners (Dutch handbook - I'm Belgian). Again ($n$) means index $n$, $x_1 = \sqrt6$, $x_{n+1} = \sqrt{6+x_n}$ Question: $$|x_{n+1} - 3| \le 1/5 \cdot |x_n - 3|...
24
votes
2answers
2k views

Sum of consecutive square roots inside a square root

$$\large\sqrt{1+\sqrt{1+2+\sqrt{1+2+3+\sqrt{1+2+3+4+\cdots}}}}$$ I saw this somewhere in the internet but, the website didn't provide me any further information. What is the sum of the equation above? ...
9
votes
2answers
396 views

Proving $(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}})(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}})=x$

How can I prove this equality? $$\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$$ if $$x>1$$
5
votes
2answers
236 views

Possible values of infinitely nested square root $n= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}…}}}$

If $$n= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}......}}}$$ Is it possible that $n$ is a integer for any $x=Z( \text{zahlen number})$.If yes .What is the value of $x$??
6
votes
1answer
432 views

Infinite square-rooting

$ \lim_{n\to\infty} {\sqrt{1+{\sqrt{2+{\sqrt{\cdots +\sqrt{n}\ }\ }\ }\ }\ \ }\ } = ? $ Either closed answer or an upper bound would help.
14
votes
1answer
445 views

Yet another nested radical

Consider $$F(x) = \sqrt{x -\sqrt{2x - \sqrt{3x - \cdots}}}$$ I believe I can prove (with some handwaving) that $F$ does converge everywhere in $\mathbb{C}$ $\Im F = 0$ for sufficiently large real $...
6
votes
2answers
213 views

Does $A_n= \sqrt{1^2+\sqrt{2^2+\sqrt{…+\sqrt{n^2}}}}$ converge?

Can we find the value of $\lim\limits_{n\to \infty} A_n$, does it converge? $$A_n= \sqrt{1^2+\sqrt{2^2+\sqrt{3^2+\sqrt{4^2+\sqrt{...+\sqrt{n^2}}}}}} $$ I tried to calculate $A_1,A_2,\cdots A_{10}$, ...
1
vote
1answer
99 views

Sufficient criterion for the convergence of $\sqrt{a_1-\sqrt{a_2-\sqrt{a_3-\cdots}}}$

In this answer, user Bill Dubuque mentioned a sufficient condition for the convergence of the infinite nested radical $\sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\cdots+\sqrt{a_n}}}}$ My question is whether there'...
0
votes
0answers
57 views

Infinitely Nested Radical with Fibonacci Coefficients

I was wondering if the following infinitely nested radical can be evaluated. $x= \sqrt{1+ \textbf{1}\sqrt{1+ \textbf{1}\sqrt{1+ \textbf{2}\sqrt{1+ \textbf{3}\sqrt{1+ \textbf{5}\sqrt{1+ \dots }}}}}} $ ...