Linked Questions

16
votes
4answers
5k views

Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$ [duplicate]

Possible Duplicate: On the sequence $x_{n+1} = \sqrt{c+x_n}$ Where does this sequence converge? $\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,...
6
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 ...
21
votes
2answers
10k views

Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? [duplicate]

Problem The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation: $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Evaluate the limit $\lim_{n\to\infty} a_n$. Solution Show that the sequence $(a_n)...
7
votes
5answers
5k views

The convergence of $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$ [duplicate]

I would like to know if this sequence converges $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$. I know this sequence is increasing monotone, but I couldn't prove it's bounded. Thanks
6
votes
2answers
5k views

Recursive square root problem [duplicate]

Give a precise meaning to evaluate the following: $$\large{\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dotsb}}}}}$$ Since I think it has a recursive structure (does it?), I reduce the equation to $$ p=\sqrt{...
3
votes
2answers
11k views

calculate the limit of this sequence $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1..}}}}$ [duplicate]

Possible Duplicate: $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ i am trying to calculate the limit of $a_n:=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+}}}}..$ ...
4
votes
2answers
12k views

Convergence and limit of a recursive sequence $x_{n+1}=\sqrt{p+x_n}$ [duplicate]

Let $p>0$ and suppose that the sequence $\{x_n\}$ is defined recursive as $$ x_1 = \sqrt{p}, \quad x_{n+1} = \sqrt{p + x_n}, $$ for all $n \in \mathbb{N}$. How can I show that $x_n$ converges, ...
3
votes
3answers
1k views

To prove a sequence is Cauchy [duplicate]

I have a sequence: $ a_{n}=\sqrt{3+ \sqrt{3 + ... \sqrt { 3} } } $ , it repeats $n$-times. and i have to prove that it is a Cauchy's sequence. So i did this: As one theorem says that every ...
3
votes
4answers
725 views

Discuss the convergence of the sequence: $a_1=1,a_{n+1}=\sqrt{2+a_n} \quad \forall n \in \mathbb{N}$ [duplicate]

Computing the first few terms $$a_1=1, a_2=\sqrt{3}=1.732....,a_3=1.9318....,a_4=1.9828...$$ I feel that $(a_n)_{n\in \mathbb{N}}$ is bounded above by 2, although I have no logical reasoning for this. ...
2
votes
3answers
493 views

Different ways of computing $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ [duplicate]

Possible Duplicate: On the sequence $x_{n+1} = \sqrt{c+x_n}$ I am wondering how many different solutions one can get to the following question: Calculate $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ ...
2
votes
4answers
3k views

Prove convergence of sequence defined recursively $a_{n+1} = \sqrt{6+a_n}$ [duplicate]

We have a sequence: $$a_1=\sqrt{6}$$ $$a_{n+1} = \sqrt{6+a_n}$$ The problem is to check convergence and then find the limit. We know that sequence converges when it is monotonic and bounded. With a ...
2
votes
4answers
424 views

Proving that the sequence $a_1 = 4$, $a_{n+1} = \sqrt{a_{n} +20}$ converges and finding its limit [duplicate]

Given a sequence $\{a_n\}$, with $n\geq1$, where $$a_{1}=4,$$ and $$a_{n+1}=\sqrt{a_{n} +20}.$$ Prove via induction that, for all $n \geq 1$, $$a_{n+1}>a_{n}.$$ Apparent Convergence ...
1
vote
2answers
880 views

Find the value of $\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$ [duplicate]

Evaluate $$\sqrt{6+\sqrt{6+\sqrt{6+\dots}}}$$ I need some help with this question because I have no idea what is going on and help would be greatly appreciate :)
2
votes
3answers
345 views

Find the value of : $\lim_{n\rightarrow \infty}\sqrt{2+\sqrt{2+\sqrt{…+\sqrt{2}}}}$ [duplicate]

This is a nice limit $$\lim_{n\rightarrow \infty}\underbrace{\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}}_{n\text{ times}}$$ and it is solved with well-known trigonometry formulas. The result is 2. The ...
4
votes
2answers
252 views

'Continued roots'' [duplicate]

Possible Duplicate: $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ Some time ago I was playing with a calculator and I found the following relation $$2 ...

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