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.


marked as duplicate by hardmath, dantopa, draks ..., Arnaldo, Daniel W. Farlow May 29 '17 at 12:45

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    $\begingroup$ As a hint, you want a recursive definition: $x_{n + 1} = \sqrt{2x_n}$ $\endgroup$ – ferson2020 Feb 28 '13 at 14:35
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    $\begingroup$ wow, thx a lot. With this hint, the inductive step becomes lots more easier $\endgroup$ – Idonknow Feb 28 '13 at 14:40
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    $\begingroup$ @Idonknow If you reached a solution, you can post it as an answer and even, after a while, if it seems correct, accept it. $\endgroup$ – Did Feb 28 '13 at 14:46
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    $\begingroup$ I misread the question, $\sqrt{2\sqrt{2\cdots}}$ is not the same as $\sqrt{2+\sqrt{2+\cdots}}$. It's a duplicate of "How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2?" instead. $\endgroup$ – Rahul May 28 '17 at 21:42

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.


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 $2^1=2$.


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.


Hint: $$2^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8}\cdots} = 2^1$$

  • $\begingroup$ (I wanted to post something different since the proofs are already posted.) $\endgroup$ – Parth Kohli Mar 6 '13 at 0:47

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_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.

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    $\begingroup$ Funny, every term is larger than 1.41. $\endgroup$ – Did Feb 28 '13 at 15:39
  • $\begingroup$ sorry I made a mistake, it's correct now(I think...) $\endgroup$ – lee Mar 3 '13 at 13:54
  • $\begingroup$ The limit is correct now but what is still wrong (or, at least, what needs some justification) is the assertion that the limsup and liminf are fixed points of the iteration. In general they are not. $\endgroup$ – Did Mar 3 '13 at 16:05
  • $\begingroup$ I know, but in this case,i.e.,it's all the $a_n$ are positive, this assertion holds. Actually I think it's the easiest way to prove the existence and calculate the limit. $\endgroup$ – lee Mar 4 '13 at 13:59
  • $\begingroup$ Even though every term of a sequence is positive, its liminf may be zero hence one could in principle have $a=2$ and $b=0$. $\endgroup$ – Did Mar 4 '13 at 16:23

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 $2$.

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    $\begingroup$ Not every bounded sequence converges. $\endgroup$ – Did Mar 6 '13 at 8:57

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