Convergence of a Recursively Defined Sequence

[Saitama]

Homework Statement

Let $x_0=2\cos(\pi/6)$ and $x_n=\sqrt{2+x_{n-1}}$, n=1,2,3,...

Find $$\lim_{n \rightarrow \infty} 2^{n+1}\cdot \sqrt{2-x_n}$$

Homework Equations

The Attempt at a Solution

I found $\displaystyle x_n=2\cos\left(\frac{\pi}{6(n+1)}\right)$

$$\Rightarrow \sqrt{2-x_n}=2\sin\left(\frac{\pi}{12(n+1)}\right)$$
The limit to be evaluated is
$$\lim_{n \rightarrow \infty} 2\cdot 2^{n+1}\cdot \sin\left(\frac{\pi}{12(n+1)}\right)$$
Evaluating it results in infinity but the given answer is $\pi/3$.

Any help is appreciated. Thanks!

 

[ehild]

Homework Statement

Let $x_0=2\cos(\pi/6)$ and $x_n=\sqrt{2+x_{n-1}}$, n=1,2,3,...

Find $$\lim_{n \rightarrow \infty} 2^{n+1}\cdot \sqrt{2-x_n}$$

Homework Equations

The Attempt at a Solution

I found $\displaystyle x_n=2\cos\left(\frac{\pi}{6(n+1)}\right)$

Are you sure? How did you arrived at it?

ehild

 

[Saitama]

Are you sure? How did you arrived at it?

ehild

I just realized that it's wrong.

The correct one is
$$x_n=2\cos\left(\frac{\pi}{6 \times 2^n} \right)$$

This gives the answer. Thank you ehild!

 

[ehild]

You see how fast you find a mistake? You do not really need us .

ehild

 

[Saitama]

You do not really need us.

No, I do. Without you people, I would never have been able to solve such problems. :)