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Let $x_{1} = -\sqrt{3}\;\;,x_{2} = -\sqrt{3+\sqrt{3}}\;\;,x_{3}=-\sqrt{3+\sqrt{3}+\sqrt{3}}\;,...... $ Then $\lim_{n\rightarrow \infty}x_{n}$

$\bf{My\; Try::}$ We can write $$x_{n} = -\underbrace{\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+.........+\sqrt{3}}}}}}_{\bf{n- times \;\sqrt{3}}}$$

So When $n\rightarrow \infty\;,$ Then $\lim_{n\rightarrow \infty}x_{n} = -\sqrt{3+\sqrt{3+\sqrt{3+..............\infty}}}$

Now how can i solve after that, help me

Thanks

juantheron
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    Look here https://math.stackexchange.com/questions/115501/sqrtc-sqrtc-sqrtc-cdots-or-the-limit-of-the-sequence-x-n1-sq –  Mar 10 '16 at 08:42

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Note that $$x_{n+1} = - \sqrt{3-x_{n}}.$$ Hence, if $x_n \to l$, one must have $$l = -\sqrt{3-l}.$$ Squaring and solving this equation gives $l = \frac{-1 \pm \sqrt{13}}{2}.$ All $x_n$ being negative, we know that the desired limit is $\frac{-1 - \sqrt{13}}{2}.$ Now you have to prove that the sequence is convergent. For instance you can first show that it is decreasing.

C. Dubussy
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