$x_{n+1} = \sqrt{3x_n}$ 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.
 A: Hint:  If it converges to $L$, you have $L=\sqrt{3L}$ which implies $L=0,3$  If $x_1 \lt 0$, you can't iterate.  Now show that if $x_i \gt 0$, $x_{i+1}-3 \lt x_i-9$ from the recursion.  That will prove it converges to $3$.
A: From $x_{n+1}=\sqrt{3x_n}$ you see that $x_n>0$ for all $n\geq0$, provided that $x_0>0$ of course. If $y_n=\log x_n$ then you have the recurrence  $y_{n+1}=\frac12y_n+\frac{\log3}2$. Multiplying by $2^{n+1}$ you get $2^{n+1}y_{n+1}-2^ny_n=2^n\log3$. Summing from $n=0$ to $k-1$  yields $2^ky_k=y_0+(2^k-1)\log3$, hence $$y_k=\frac{y_0+(2^k-1)\log3}{2^k}\,.$$ Taking exponential you obtain an explicit formula for $x_k$:
$$x_k=\exp\Bigl(\frac1{2^k}\log x_0\Bigr)\,\exp\biggl(\Bigl(1-\frac1{2^k}\Bigr)\log3\biggr)=x_0^{1/2^k}\,3^{1-1/2^k}$$
The rest is left to you (perhaps you will need L'Hôpital's rule).
A: 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 $x_0 \in I$.
A: 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 $x=3$.
A: A different way to approach this problem is to note that $x_1 < 0$ makes it undefined at $x_1=0$ is a fixed point. For a positive $x_i$, we produce $x_{i+1}$ by taking a geometric average of $x_i$ with 3, so repeated application will converge to 3 for any finite positive $x_1$...
