Linked Questions

66
votes
4answers
12k views

$\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, 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 ...
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
5answers
2k views

Why is $\sqrt{2\sqrt{2\sqrt{2\cdots}}} = 2$? [duplicate]

Why is $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}}}$ equal to 2? Does this work for other numbers?
7
votes
2answers
21k 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 ...
15
votes
4answers
2k views

Nested Radicals: $\sqrt{a+\sqrt{2a+\sqrt{3a+\ldots}}}$

Let $a>0$ . How we can find the limit of : $$\sqrt{a+\sqrt{2a+\sqrt{3a+\ldots}}}$$ Thanks in advance for your help
20
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 ...
8
votes
3answers
1k 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|...
4
votes
4answers
436 views

Request for example of a value of limit of a sequence where sequence does not converge

In Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ Timothy Wagner gave a correct answer that was questioned for not having shown that the limit exists in the first place. My question ...
3
votes
2answers
4k views

Repeating Square Root Simplification

Alright, so I have a question on a little open-book challenge-test thingy that deals with repeating square roots, in a form as follows... $\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ Repeated 2012 times (...
5
votes
2answers
116 views

Convergent Sequence Terminology

What is the following sequence classified as? I don't want to make anybody solve it, I just need to know where to begin looking to solve it. $$\alpha_1 = \sqrt{20}$$ $$\alpha_{n+1} = \sqrt{20 + \...
4
votes
2answers
134 views

How to compute the following formulas? [duplicate]

$\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$ $\dots\sqrt{2+\sqrt{2+\sqrt{2}}}$ Why they are different?
7
votes
1answer
389 views

Sylow 2-subgroup of GL(2,R)

I was trying to work out the Sylow p-subgroups for general linear groups over arbitrary fields, and was running into some trouble with non-algebraically closed fields. The real numbers, R, were ...
0
votes
1answer
51 views

Proof by induction: Let $a_0=3$ and $a_{n+1}=\sqrt{a_n+7}$ if $n>0$, Prove: $3<a_n<4$

Let $a_0=3$ and $a_{n+1}=\sqrt{a_n+7}$ if $n>0$ Prove: $3<a_n<4$ At first I was quite surprised it's actually true for the base cases: $n=0$, $a_1=\sqrt{3+7}=\sqrt{10}$ $n=1$, $a_2=...