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.

Collected in the board: Limit

Steven Zheng posted 3 days ago

Answer 1

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\to \infty} a_n=\lim\limits_{n\to \infty} 2\cos\dfrac{\pi}{2^{n+1}}=2

Steven Zheng posted 2 days ago

Answer 2

Steven Zheng Steven Zheng posted the answer 9 hours ago

This answer is set private or in draft.

Scroll to Top