Prove $\lim\limits_{n\to\infty}\sqrt{2+\sqrt{2+...+\sqrt{2}}}=2$
How to evaluate this limit?
Asked
Active
Viewed 109 times
0
-
3Hint: show that the sequence is monotonic and bounded from above. It will follow that the limit exists. Then use properties of limits to show it must be $2$. – Wojowu Sep 23 '15 at 14:26
-
1look at it as the sequence of numbers ${x_n}$ where $x_1 = \sqrt{2}$ and $x_{n+1}=\sqrt{2+x_n}$. Then do it as Wojowu suggests – R_D Sep 23 '15 at 14:28
-
2possible duplicate of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ – Hans Lundmark Sep 23 '15 at 15:04
2 Answers
4
The standard idea would be to define a sequence $a_n$ as follows: $a_0=\sqrt{2}$ and $a_{n+1}=\sqrt{2+a_n}$ for all $n \ge 0$. Then you want to evaluate $\lim_{n \to \infty} a_n$. One way to prove convergence is to prove that $a_n$ is bounded above and mononotically increasing. Both of them should not be too hard. Then, if you know that it converges to the limit $L$ you find $L=\sqrt{2+L}$ and then can solve for $L$. Can you complete this proof on your own?
Tintarn
- 2,931
3
It is the limit of the sequence defined by $u_0=0$ and $\;u_{n+1}=\sqrt{2+u_n}$. As the function $f(x)=\sqrt{2+x}$, is continuous the limit $\ell$ is a nonnegative fixed point of the function, i.e. is satisfies the equation: $$\ell=\sqrt{\ell+2}\iff \ell^2-\ell-2=0\enspace\text{and}\enspace \ell\ge 0 $$ Now this equation has two roots, $2$ and$-1$. You eliminate the negative root.
As noticed by @Thomas Andrews, this supposes you've proved there is a limit first. As the function maps $[-2,2]$ to $[0,2]$, and $f(x)>x$ on this interval, we have an increasing, bounded from above sequence, which therefore converges to a limit.
Bernard
- 175,478
-
-
@Thomas Andrews: Of course, but the question, as formulated by the O.P., was not about how to prove the sequence converges. I'll add this remark. – Bernard Sep 23 '15 at 14:43
-
"How to evaluate this limit" does not imply "if this limit exists," so, no, your are wrong. – Thomas Andrews Sep 23 '15 at 16:35
-
For me, it meant that he/she had solved the question of the existence of the limit. Otherwise the question would have been something like ‘ how to show there's a limit’. At least that's the way I'd formulate it. – Bernard Sep 23 '15 at 16:41