Showing that a Sequence is defined recursively

Question

How can I show that the sequence is defined recursively?

Show that the recursively defined sequence $(x_n)_{n\in\mathbb{N}}$ with $$x_1=1, \qquad\qquad x_{n+1}=\sqrt{6+x_n}$$ converges and determine its limit

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Answer 1

The sequence is defined recursively, with initial value $x_1=1$ and with recursive relation $x_{n+1}=\sqrt{6+x_n}$. Your task is to prove that it converges and to determine its limit.

Convergence: There is a theorem that says that if a sequence is monotonically increasing and bounded from above, then it has a limit. This suggests the following:

1. Show that the sequence is monotonically increasing.

2. Show that the sequence is bounded from above.

Computing the limit: use the fact that as $n$ tends to infinity, both $x_{n+1}$ and $x_n$ are roughly equal to the limit $L$.

3. Set $x_{n+1}=L$ and $x_n=L$, and extract the value of $L$.

Answer 2

$\sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{7}}}}<x$

If we go by induction :

$\sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{7}}}}=U_n<x$

$(U_{n+1})²-6=U_n<x$

$U_{n+1}<\sqrt{x+6}$

$\sqrt{x+6}=x$

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• Now it is left to you calculating $x$ and $U_0$ testing part.