Let $X_n \,\,(n∈N)$ be a sequence defined by $X_1 = 1, X_{n+1} = \sqrt{2X_n}$ and $n ≥ 1$. Show that $X_n$ is convergent.
So I guess ${X_n}$ goes to a number like $$\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\dots}}}}}.$$ But i'm not even sure if this makes $X_n$ convergent. And therefore can't prove it.
First time I'm asking a question here. Sorry if i did something wrong.
Step I:
Use induction to show that each term is less than $2$. Inductive Step is $$X_k<2\implies X_{k+1}=\sqrt{2X_k}<\sqrt{2\times 2}=2$$
Step II:
Use induction to show that given sequence is increasing. Inductive Step is $$X_{k}>X_{k-1}\implies\sqrt{2X_k}>\sqrt{2X_{k-1}}\implies X_{k+1}>X_k$$
Since it is a monotonic bounded sequence, hence convergent.
Well, if $x_n\to L$ then $\sqrt{2x_n}\to\sqrt{2l}$, so we must have $L=\sqrt{2L}$, or $L=2$.
So showing that $x_n$ converges is the same as showing that $x_n\to2$. Which motivates the following: It follows by induction that $x_n<2$. Hence $x_{n+1}^2=2x_n>x_n^2$. So the sequence is monotone and bounded, hence convergent.
To show that it's convergent, we can show that's it's monotone increasing and bounded above (by $2$). To show that it's monotone increasing, we need to show that $X_n < X_{n+1} $ for $n\geq 1$. We can do so by induction. When $n=1, 1=X_n < X_{n+1}=\sqrt{2},$ so the base case holds. Assume the result holds for some $k\in\mathbb{N}, k < n$. Then $X_k < X_{k+1}$. Since $X_{k+2} =\sqrt{2X_{k+1}}>\sqrt{2X_k}$ by hypothesis, the result holds by induction.
Now, to show that it's bounded above by $2$, we see that the base case holds as $X_1 = 1 <2$. Assume $X_k < 2$ for some $k\in\mathbb{N}, k < n$. Then $X_{k+1} =\sqrt{2X_k} < \sqrt{2\cdot 2} = 2$, so the result holds by induction.
Hence by MCT the sequence converges.
