Show that the sequence $\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$ converges and find its limit. Show that the sequence $$\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$$ converges and find its limit. I put the sequence in this form , $(x_n)$ where $$\large x_n=2^{\Large\sum_{k=1}^{n}\left(\frac{1}{2^k}\right)}$$ I want to use Monotonic Convergence Theorem to show. But I stuck at proving the sequence is bounded above by 2. I manage to prove the sequence is an increasing sequence. Anyone can guide me ? I try to use induction to prove but i stuck at inductive step.
 A: If you already proved the exponential form of $x_n$, then, it is easy, as $t\mapsto 2^t$ is continuous, and the exponent
$$\sum_{k=1}^n \frac1{2^k} \  = 1-\frac1{2^{n+1}} \ \ \overset{n\to\infty}\longrightarrow \ 1 \ ,$$
so your limit is $2^1=2$.
A: We have the recursive definition $a_1=\sqrt{2},a_{n+1}=\sqrt{2a_n}.$ (Hopefully, that's clear.)
You've shown that the $a_n$ form an increasing sequence, and you know that $a_1<2.$ For the induction step, we want to show that if $a_n<2,$ then $a_{n+1}<2.$ Note that all the $a_n$ are positive (since $a_1$ is and the sequence is increasing), and note that for $0\leq x<y$ we have $\sqrt{x}<\sqrt{y}$. Thus, if $a_n<2$, then $2a_n<4$, and so $$a_{n+1}=\sqrt{2a_n}<\sqrt{4}=2,$$ as desired.
In general, if you're trying to prove something like this, you'll want to be able to rewrite $a_{n+1}$ in terms of $a_n$ to allow the induction step to work.
A: Hint: $$2^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8}\cdots} = 2^1$$
A: Let $x^* = \lim_{n\to\infty} a_n$ 
Notice that $a_n = \sqrt{2a_{n-1}}$.
Hence $\lim_{n\to\infty} a_n = \lim_{n\to\infty} \sqrt{2a_{n-1}}$ $\Rightarrow $
$$
x^* = \sqrt{2x^*} \Longrightarrow\\ (x^*)^2 - 2x^* = 0 \Longrightarrow\\ x^* = 2 \;\;\;\text{ or } \;\;\; x^* = 0 
$$
So, once you've established that the sequence is non-decreasing,
$$
x^* = \lim_{n\to\infty} a_n = 2
$$
A: $a_n^2=$2a_n+1,suppose a is the upper limit of $a_n$, b is the lower limit of $a_n$, then you have $a^2=2a$（sorry I made a mistake here, thx for pointing out.）, and $b^2=2b$, and a,b>0, thus a=b=2. So you can conclude the limit exists and equals to 2.
A: The geometric series $\sum_{k=1}^{\infty}(1/2)^k = 1$ .
Hence $x_n = 2^{\sum_{k=1}^n (1/2)^k} \le 2$. 
Hence, The sequence is bounded above by $2$ and thus the series converges as it is increasing. Now you can use the hint given by ferson2020 to see that the limit $L$ has the property $L = \sqrt{2L}$ and hence $L = 2$. 
Added: I am not sure if you proved the formula 
$$x_n = 2^{\sum_{k=1}^n (1/2)^k}$$ so I give a proof here. We proceed with induction on n. The base case $n = 1$ holds by definition. Now assume it holds for $x_n$, we show that it is holds for $x_{n+1}$. 
We have $x_n = \sqrt{2x_n}$. 
So by the inductive hypothesis: $$x_{n+1} = \sqrt{2 \cdot 2^{\sum_{k=1}^n (1/2)^k}} = 2^{({\sum_{k=1}^n (1/2)^k} +1)/2}$$ 
Now ,
$ S = \sum_{k=1}^n (1/2)^k = 1/2 + 1/4 + \ldots 1/2^n$ 
and if we multiply this with $1/2$ we get $1/4 + 1/8 + \ldots 1/2^{n+1}$.
So $$S/2 = \sum_{k=2}^{n+1}(1/2)^k$$ 
Then $$S/2 + 1/2 = \sum_{k=1}^{n+1}(1/2)^k$$ 
So we have $x_{n+1} = 2^{(S +1)/2}= 2^{\sum_{k=1}^{n+1}(1/2)^k}$ this completes the proof.
A: The recursive relation can be defined as
$$
x_n = \sqrt{2x_{n-1}}
$$
with $x_0 = 1$. It's clear by induction that 0 < $x_n < 2$, hence the sequence converges. The limit must satisfy
$$
x_{\infty} = \sqrt{2x_\infty}
$$
and so the limit is $2$. 
