# 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 question is: can be, this limit, solved in other ways?

Thanks.

## marked as duplicate by Najib Idrissi, Micah, Asaf Karagila♦, Jean-Claude Arbaut, Fly by NightAug 23 '14 at 20:45

If $x_0=\sqrt 2$ and $x_{n+1}=\sqrt{2+x_n}$, then you can prove by induction that $x_{n}<2$ and $x_n$ is increasing. So a limit must exist.

But any limit must be a positive $x$ satisfying $x = \sqrt{2+x}$. So $x=2$.

Seeing as there is no $n$ in your expression, having $\lim_{n \to \infty}$ doesn't really make sense. I would just express it as $\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}$

Just suppose there is some number $A$ such that $A = \sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}$. then we can square and subtract two from both sides and get $A^2 -2 = \sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}$. But the right-hand side is the same in both of these, so we have $A^2 - 2 = A$. Solve this quadratic for $A$ and we see that in fact $A = 2$

• But you need to prove that such a limit exists. For example, $1,3,9,\dots$ would have limit $0$, since any limit must have $3A=A$, that means the limit must be $0$. If it exists. :) – Thomas Andrews Aug 23 '14 at 19:51
• The expression $n$ is the number of terms in the square root. There is no infinite expression of that form that "terminates." – Thomas Andrews Aug 23 '14 at 19:53
• @ThomasAndrews However, mapierce271 is right in saying there is no $n$ in the expression, thus the question is not asked correctly. Usually, there is at list a line to tell what you write. Without that, there could be $n^2$ square roots, or $n\ \mathrm{mod} \ 10$, or whatever. – Jean-Claude Arbaut Aug 23 '14 at 20:01
• The expression is meaningless if it implies an infinite number of terms. It is true, OP didn't say what he meant, but this answer implies it makes sense to write $A = \sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}$ as if there were such an expression with infinite number of terms. The terminating $\sqrt{2}$ means the definition doesn't makes sense for infinite terms. @Jean-ClaudeArbaut – Thomas Andrews Aug 23 '14 at 20:10
• The formula now has a notation indicating that there are $n$ nested square roots. Hopefully that resolves the "number of terms" concern. – David K Aug 23 '14 at 20:16

If the limit exists, the recursive function is very simply: $$n=\sqrt{2+n}$$

Hence:

$$n^2-n-2=0 \\ (n-2)(n+1)=0 \\n=2$$

$n$ is indeed $2$.

• That assumes a limit exists, of course... – Thomas Andrews Aug 23 '14 at 19:49
• @ThomasAndrews I assume the OP is given that the limit exists. – Shahar Aug 23 '14 at 19:50