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

Suppose \{a_n\} has limit,

let \lim\limits_{n\to \infty} a_n=C

Then \lim\limits_{n\to \infty} a_{n+1}=C

Given the formula of the recursive sequence

a_{n+1}=\sqrt{2+a_n}

Take the limit

\lim\limits_{n\to \infty} a_{n+1}=\sqrt{2+\lim\limits_{n\to \infty} a_n}

C=\sqrt{2+C} >0

Taking the square and rearranging the terms gives

C^2-C-2=0

Solve the quadratic equation

C=2 or C=-1 (canceled )

Now we have found the sequence has a limit and it’s value is 2.