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$$a_1=1,a_{n+1}=\sqrt{6+a_n}, n \in \mathbb{N}$$

Show that the limit $\lim\limits_{n \to +\infty} a_n$ exists and find it.

I know to prove that the sequence is bounded and monotonous but I still don't know how to find the limit.

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One can show by induction that $a_n \leqslant 3$ and monotonically increasing, which means that the limit exists. Suppose the limit is $L$. Then,

$$L = \sqrt{6+L} \Rightarrow L^2 - L - 6=0 \Rightarrow L = 3$$

So, $\lim\limits_{n \to +\infty} a_n = L = 3$.

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  • $\begingroup$ Thanks but why do we choose to '3 value' for upper bound? $\endgroup$ – bopele Mar 31 '17 at 16:34
  • $\begingroup$ @bopele At the limit $a_{n+1} = a_n = L$. Solving the equation $L^2 - L - 6 = 0$ gives you three. Knowing that show $a_n \leq 3$ by induction. $\endgroup$ – Χpẘ Mar 31 '17 at 16:43
  • $\begingroup$ You could take both roots depending on the behavior of the function. The lesser root gives bounded below and the upper root gives bounded above. Since you are showing monotonically increasing you have to work with bounded above because by the bounded monotonic convergence theorem. $\endgroup$ – user349557 Mar 31 '17 at 19:40

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