TITLE: $x_{n+1} = \sqrt{3x_n}$ converges for $x_1 = 1,x_1 = 27$ Proof
QUESTION [6 upvotes]: Prove $x_{n+1} = \sqrt{3x_n}$ converges for $x_1 = 1,x_1 = 27$. (Separate problems for $x_1 = 1$ and $x_1 = 27$.)
EDIT: Took out bad algebra.

REPLY [2 votes]: You can solve both problems at the same time.
Let $f(x)  = \sqrt{3x}$. Note that $f'(x) = \frac{\sqrt{3}}{2\sqrt{x}}$, and if $x \in I=[1,27]$, then $0 < f'(x) < 1$. (So, in particular, $|f'(x)| \leq \lambda <1$, for $x\in I$.) Furthermore, note that  $3$ is the only fixed point of $f$  in $I$.
The mean value theorem gives $|x_{n+1}-3| \leq \lambda|x_n-3|$, and since $0<f'(x)$, we have that the sign of $x_{n+1}-3$ and $x_{n}-3$ are the same. Hence $f(I) \subset I$. Since $f$ is a contraction on $I$, it follows that $x_n \to 3$, as long as $x_0 \in I$.

REPLY [1 votes]: There's no need to prove this using induction as you can find the value that it converges to!
Begin by letting $x_1=a$ and finding the first few terms, as follows:
\begin{align*}
x_1 &= a \\
x_2 &= \sqrt{3}\cdot\sqrt{a} = \sqrt{3a} \\
x_3 &= \sqrt{3}\cdot\sqrt{\sqrt{3a}} = \sqrt{3\sqrt{3a}}\\
x_4 &= \sqrt{3}\cdot\sqrt{\sqrt{3\sqrt{3a}}} = \sqrt{3\sqrt{3\sqrt{3a}}}\\
\end{align*}
As you can see, there is an evident pattern in the terms. It can be represented by $x=\sqrt{3\sqrt{3\sqrt{3\sqrt{3...}}}}$.
At $x_1=1$, $x=\sqrt{3x}$, which gives an obvious solution of $x=3$.
At $x_1=27$, the equation simplifies to the same $x=\sqrt{3x}$ because the infinite square root makes the 27 negligible, which gives an obvious solution of $x=3$.
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