Where does this sequence converge?
$\sqrt{7},\sqrt{7+\sqrt{7}},\sqrt{7+\sqrt{7+\sqrt{7}}}$,...
Answer
For a proof of convergence,
Define the sequence as
$\displaystyle x_{0} = 0$
$\displaystyle x_{n+1} =\sqrt{7 + x_n}$
Note that $\displaystyle x_n \geq 0 \ \ \forall n$.
Notice that $\displaystyle x^2 - x - 7 = (x-a)(x-b)$ where $\displaystyle a \lt 0$ and $\displaystyle b \gt 0$.
We claim the following:
i) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \lt b$
ii) $\displaystyle x_n \lt b \Longrightarrow x_{n+1} \gt x_n$
For a proof of i)
We have that
$\displaystyle x_n \lt b = b^2 - 7$ and so $x_n +7 \lt b^2$ and thus by taking square roots $x_{n+1} \lt b$
For a proof of ii)
We have that
$\displaystyle (x_{n+1})^2 - (x_n)^2 = -(x^2_n - x_n -7) = -(x_n-a)(x_n-b) \gt 0$ if $x_n \lt b$.
Thus $\displaystyle \{x_{n}\}$ is monotonically increasing and bounded above and so is convergent.
By setting $L = \sqrt{7+L}$, we can easily see that the limit is $\displaystyle b = \dfrac{1 + \sqrt{29}}{2}$
In fact, we can show that the convergence is linear.
$\displaystyle \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{b^2 - (7+x_n)}{(b+\sqrt{7+x_n})(b-x_n)} = \dfrac{1}{b + x_{n+1}}$
Thus $\displaystyle \lim_{n\to \infty} \dfrac{b-x_{n+1}}{b-x_n} = \dfrac{1}{2b}$.
We can also show something a bit stronger:
Let $\displaystyle t_n = b - x_n$.
The we have shown above that $\displaystyle t_n \gt 0$ and $\displaystyle t_n \lt b^2$
We have that
$\displaystyle b - t_{n+1} = \sqrt{7 + b - t_n} = \sqrt{b^2 - t_n}$
Dividing by $\displaystyle b$ throughout we get
$\displaystyle 1 - \dfrac{t_{n+1}}{b} = \sqrt{1 - \dfrac{t_n}{b^2}}$
Using $\displaystyle 1 - \dfrac{x}{2} \gt \sqrt{1-x} \gt 1 - x \ \ 0 \lt x \lt 1$ we have that
$\displaystyle 1 - \dfrac{t_n}{2b^2} \geq 1 - \dfrac{t_{n+1}}{b} \geq 1 - \dfrac{t_n}{b^2}$
And so
$\displaystyle \dfrac{t_n}{2b} \leq t_{n+1} \leq \dfrac{t_n}{b}$
This gives us that $\displaystyle b - \dfrac{b}{b^n} \leq x_n \leq b - \dfrac{b}{(2b)^n}$
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