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Let $a_{n+1}=\sqrt{2+a_n}, a_1=\sqrt{3}$. Show that it is convergent.

I know that this is a classic nested square root sequence but how do I prove it's convergence? To know it's limit I can just take the limit on both sides and find $L$, or rewrite $a_n=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}$

Nosrati
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C. Cristi
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  • Bounded and increasing. –  Nov 25 '18 at 09:14
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    $$a_n<2\implies a_n+2<4\implies a_{n+1}<2.$$

    $$\frac12<a_n<2\implies a_n^2-a_n-2=\left(a_n-\frac12\right)^2-\frac94\le0\implies a_n\le\sqrt{a_n+2}=a_{n+1}.$$

    Hence the sequence is bounded and monotonous.

     –  Nov 25 '18 at 09:28

1 Answers1

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Define $$b_n=a_n-2$$therefore $$b_{n+1}=a_{n+1}-2=\sqrt{a_n+2}-2={a_n-2\over 2+\sqrt{a_n+2}}={b_n\over 2+\sqrt {a_n+2}}$$therefore $$|b_{n+1}|=|{b_n\over 2+\sqrt {a_n+2}}|\le |{b_n \over 2}|\le \cdots \le {|b_1|\over 2^n}$$therefore $b_n\to 0$ when $n\to \infty$ which means that $a_n\to 2$

Mostafa Ayaz
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