1. convergent sequence

Suppose $x_1=\sqrt{2}$ and $x_{n+1}=\sqrt{2x_n}$ for $n\geq 1$. Show that the sequence converges.

2. Originally Posted by Kat-M
Suppose $x_1=\sqrt{2}$ and $x_{n+1}=\sqrt{2x_n}$ for $n\geq 1$. Show that the sequence converges.
$x_1 < x_2 = \sqrt {2x_1 } = \sqrt {2\sqrt 2 } < 2$

$\begin{gathered}
x_{N - 1} < x_N < 2 \hfill \\
2x_{N - 1} < 2x_N < 4 \hfill \\
\sqrt {2x_{N - 1} } < \sqrt {2x_N } < \sqrt 4 \hfill \\
x_N < x_{N + 1} < 2 \hfill \\
\end{gathered}$

So it increasing bounded above. Thus $\left( {x_N } \right) \to L\;\& \,L = \sqrt {2L}$