<s>[INST] $x_{n+1} = \sqrt{3x_n}$ converges for $x_1 = 1,x_1 = 27$ Proof 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.
 [/INST] 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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