If $$n\in N^*$$ such that $$a_1=\sqrt{2} $$, $$a_{n+1}=\sqrt{2+a_n}$$, show that $$a_n$$ has limit and find the value of the the limit.

Question

If $$n\in N^*$$ such that $$a_1=\sqrt{2}  $$,  $$a_{n+1}=\sqrt{2+a_n}$$, show that $$a_n$$ has limit and find the value of the the limit.

Uniteasy Admin · Accepted Answer

$$a_1=\sqrt{2}=2\cos\dfrac{\pi}{4}  $$
$$a_2=\sqrt{2+a_1} =\sqrt{2+2\cos\dfrac{\pi}{4} } $$
$$\cos\dfrac{ \alpha}{2} = \sqrt{\dfrac{1+\cos \alpha }{2} }$$
Using the half angle identity 
$$\cos^2\dfrac{\alpha }{2}=\dfrac{1+\cos\alpha }{2}  $$
$$a_2=\sqrt{2^2\cdotp \cos^2\dfrac{\pi}{8}  } =2\cos\dfrac{\pi}{8}  $$
Similarly,
$$a_3=2\cos\dfrac{\pi}{16} $$
$$\dots$$
$$a_n= 2\cos\dfrac{\pi}{2^{n+1}} $$
$$\lim\limits_{n	o \infty} a_n=\lim\limits_{n	o \infty} 2\cos\dfrac{\pi}{2^{n+1}}=2$$

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