How prove that $(a_{n})_{n \in \mathbb{N}}$ is convergent with $a_{0}=1$, $a_{n+1}= \sqrt{2a_{n}}$?

Prove that $(a_{n})_{n \in \mathbb{N}}$ is convergent with $a_{0}=1$, $a_{n+1}= \sqrt{2a_{n}}$ and calculate its limit.

You got the info that the square-root-function is strictly monotonic increasing and also continuous.

I'm very confused how this could be solved because there isn't really a sequence given here. It seems to be a recursive one and I also don't know how to form $a_{n+1}$.

Could I form it to $a_{n}+ a_{1}$ ?(I don't care if it's useful here but I'd like to know if this would be correct at first.)

I have started by forming to $a_{n}$:

$$2a_{n} = (a_{n+1})^{2}$$

$$a_{n}=\frac{(a_{n+1})^{2}}{2}$$

Then I also formed to $a_{1}:$

$$a_{n}+ a_{1} = \sqrt{2a_{n}}$$

$$a_{1} = \sqrt{2a_{n}} - a_{n}$$

But it seems like this won't lead to anything.. Also I don't know what we can do with the given info, $a_{0} = 1$, the continuity and monotonic increasing...

The most basic, but still readable proof (in my opinion) is by using the following theorem:

A monotonically increasing, bounded sequence of real numbers is convergent.

So we need to prove that $a_n$ is increasing and bounded. Specifically, I will prove that $1\leq a_n \leq a_{n+1} \leq 2$ for all $n$ using induction.

Base case: The inequality clearly holds for $n = 0$, since $1 \leq 1 \leq \sqrt 2 \leq 2$.

Induction step: Assume the inequality holds for $n = k$, i.e. that $1\leq a_k \leq a_{k+1}\leq 2$. We want to show $$ 1 \leq a_{k+1} \leq a_{k+2} \leq 2 $$ The first inequality follows from the induction hypothesis and transitivity of $\leq$ (we assumed $1 \leq a_k$ and $a_k \leq a_{k+1}$). For the second inequality, we have $$ a_{k+1} = \sqrt{a_{k+1}\cdot a_{k+1}} \leq \sqrt{2a_{k+1}} = a_{k+2} $$ where we used $ a_{k+1}\leq 2$, as well as the monotonically increasing nature of the square root.

For the last inequality, we have $$ a_{k+2} = \sqrt{2\cdot a_{k+1}} \leq \sqrt{2\cdot 2} = 2 $$ which concludes the proof that $1 \leq a_{k+1} \leq a_{k+2} \leq 2$.

By induction $$a_{n+1}=\sqrt{2a_n}=\sqrt{2\cdot\sqrt{2a_{n-1}} }=2^{1/2}2^{1/4}a_{n-1}^{1/4}=...=a_0\prod_{i=1}^{n+1}2^{\frac{1}{2^{i}}}.$$

As robjohn suggest, define $b_{n+1}=\log(a_{n+1})$ and conclude.