# 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?

## marked as duplicate by YuiTo Cheng, Paul Frost, cmk, max_zorn, ThorWittichJul 11 at 22:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• The second thing must be $\lim_{n\to \infty} x_n$ where $x_0=0$ and $x_{n+1}=\sqrt{x_n+2}$. But expressions like the first one always confuse me a bit. It's obviously supposed to be some positive solution to $\sqrt{2+x}=x$, but it feels a bit ambiguous – Myself Mar 9 '11 at 5:45
• A proof of convergence can be found here: math.stackexchange.com/questions/11945/…. With 7 replaced by 2, the proof basically carries over. – Aryabhata Mar 9 '11 at 6:05
• @Rasmus Do you know how to rigorously 'define' that first thing? To me it seems both limiting points would satisfy the same equation $x=\sqrt{x+2}$. – Myself Mar 9 '11 at 7:10
• @user7992 Answer: They are not, since both numbers equal $2$. – Did Mar 9 '11 at 7:35
• @Myself: I deleted my previous comment because I misread the question. – Rasmus Mar 9 '11 at 10:50

## 2 Answers

Suppose that the first converges to some value $x$. Because the whole expression is identical to the first inner radical, $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}=x=\sqrt{2+x}$ and solving for $x$ gives $x=2$. Of course, I haven't justified that it converges to some value.

The second can be thought of as starting with $\sqrt{2}$ and repeatedly applying the function $f(x)=\sqrt{2+x}$. Trying this numerically suggests that the values converge to 2. Solveing $f(x)=x$ shows that $2$ is a fixed point of that function.

Looking at the second expression is actually how I'd justify (though it is perhaps not a rigorous proof) that the first expression converges.

Define a sequence $(x_n)_{n\geq1}$ so that $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2+x_n}$. Then the second formula you give can be said to colorfully describe the limit $\lim_{n\to\infty}x_n$. If we suppose the limit $L$ does exist (and it is not difficult to show it does exist!), then since for all $n$ we have $x_{n+1}^2=2+x_n$, passing to the limit we see that $L^2=2+L$. The polynomial $x^2-x-2$ has two roots, $-1$ and $2$: since all the $x_n$ are positive, the only possible value for $L$ is $2$.

Can you do the other one?

• Thank you very much. But I still can't solve the first one, because I can't write the recursive formula for the first one as you gave for the second one. My teacher told me that they were different. But from the others' replies, it seems they are the same. I'm confused. – Timothy Mar 9 '11 at 6:43
• You could ask your teacher :) – Mariano Suárez-Álvarez Mar 9 '11 at 6:52
• I wonder what motivated the downvote... – Mariano Suárez-Álvarez Mar 9 '11 at 22:29