# How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2?

Prove that the sequence $$\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \$$ converges to $$2$$.

My attempt

I proved that the sequence is increasing and bounded by $$2$$, can anyone help me show that the sequence converges to $$2$$? Thanks for your help.

• you actually know that it converges to 2? – HipsterMathematician Sep 21 '12 at 18:19
• No,That's all I need to prove – Miguel Mora Luna Sep 21 '12 at 18:21
• @kingW3: Oops, I completely misread the question. I've retracted my close vote. – user856 May 28 '17 at 16:53
• @Rahul Just wanted to point that out :P (as a side note I voted to reopen the other two questions so they can be correctly closed as dupes) – kingW3 May 28 '17 at 22:02
• A cool thing about this sequence: Let's first introduce a notation: $t_0=0,\ t_{n+1}=\sqrt{2+t_n}.$ We know that $t_n \to 2.$ Then $2-t_n$ obviously converges to $0$. The rate of convergence is such that $(2-t_{n+1})/(2-t_n) \to 1/4.$ Therefore $q_n = 4^n (2-t_n)$ converges to a finite (non-zero) number. Instead of trying to recognize that number, study the values of $p_n = 2^{n+1} \sqrt{2-t_n},$ the limit of which you will certainly recognize. – md2perpe Mar 21 at 11:45

Another prove:

Notice that: $a_1 = 2^{1/2},\ a_2 = 2^{3/4},\ a_3 = 2^{7/8}$, and so on, thus, $$a_n = 2^{(2^n-1)/2^n} = 2^{1-1/2^n}$$

Taking limit as $n$ tends to infinity, we have that $$\lim_{n \rightarrow \infty} a_n = 2^1 = 2$$

We have

$$\sqrt{2}=2^{\frac{1}{2}}.$$ Exciting, no? We also have $$\sqrt{2\sqrt{2}}=2^{\frac{1}{2}+\frac{1}{4}},$$ and $$\sqrt{2\sqrt{2\sqrt{2}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}.$$ For the next term, we take the previous term, multiply by $2$, getting exponent of $2$ equal to $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$, then take the square root, getting exponent of $2$ equal to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$. The pattern continues, since the "next term" is always obtained by the same process.

It is well-known that the series $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$ converges to $1$. Since the function $f(x)=2^x$ is continuous, our original sequence converges to $2^1$.

• Awesome answer. – ThomasMcLeod Sep 22 '12 at 3:20
• +1, Great answer. indeed. :-) Not even god knows why this was not accepted or rather is still not accepted. – SchrodingersCat Dec 15 '15 at 8:57
• Awesome. A no-nonsense, soul-satisfying kind of answer. Love it. – Mallory May 12 '16 at 4:51

Put

$$a_n:=\underbrace{\sqrt{2\sqrt{2...\sqrt 2}}}_{n\text{ square roots}}\Longrightarrow a_n=\sqrt{2a_{n-1}}$$

Since you've already proved the sequence $\,\{a_n\}\,$ converges, assume its limit is $\,x\,$ , so

$$x=\lim_{n\to\infty}a_n=\lim\sqrt{2a_{n-1}}=\sqrt{2x}\Longrightarrow x^2=2x$$

and since it's trivial that $\,x\neq 0\,$ then you can cancel above and get what you want.

Hint : let $y = \sqrt{2.{y}}$ , Now solve for y.

• Why did you make that y? – Miguel Mora Luna Sep 21 '12 at 18:24
• I also planned to do that but not to give justification – Miguel Mora Luna Sep 21 '12 at 18:25
• @MiguelMoraLuna: quartz is computing the steady-state of the iteration $x_{k+1} = \sqrt{2 x_k}$. – Rod Carvalho Sep 21 '12 at 19:11
• You'll also need to prove that x isn't 0 for this approach to work. – David Bandel Jul 13 '16 at 3:43

Consider the following discrete-time dynamical system (or "infinite state machine", if you prefer)

$$x_{k+1} = \displaystyle\sqrt{2 x_k}$$

where $x_0 = \sqrt{2}$ is the initial condition. Let us introduce function $f (x) := \sqrt{2 x}$ so that the state-transition equation can be written in the form $x_{k+1} = f (x_k)$. You want to prove that

$$x_{\infty} := \displaystyle\lim_{k \rightarrow \infty} x_k = 2$$

Does $x_k$ converge to $2$? Here's a short numerical experiment in Haskell (using GHCi):

*Main> let f x = sqrt (2 * x)
*Main> let x0 = sqrt (2)
*Main> let xs = iterate f x0
*Main> take 10 xs
[1.4142135623730951,1.6817928305074292,1.8340080864093424,1.9152065613971472,1.9571441241754002,1.978456026387951,1.9891988469672663,1.9945921121709402,1.9972942257819404,1.9986466550053015]


Looks like $x_k$ does indeed converge to $2$. Let's now try to prove it. I introduce a new symbol

$$\tilde{x}_{k} := \ln (x_k)$$

which allows us to rewrite the state-transition equation $x_{k+1} = \displaystyle\sqrt{2 x_k}$ in the following form

$$\tilde{x}_{k+1} = \frac{1}{2} \tilde{x}_k + \ln(\sqrt{2})$$

which is a scalar linear dynamical system of the form $s_{k+1} = a s_k + b$, whose general solution is

$$s_ k = a^k s_0 + \displaystyle\sum_{i=0}^{k-1} a^i b$$

Therefore, we obtain the following general solution

$$\tilde{x}_k = \left(\frac{1}{2}\right)^k \tilde{x}_0 + \displaystyle\sum_{i=0}^{k-1} \left(\frac{1}{2}\right)^i \ln(\sqrt{2})$$

and, taking the limit

$$\tilde{x}_{\infty} := \displaystyle\lim_{k \rightarrow \infty} \tilde{x}_k = \displaystyle\lim_{k \rightarrow \infty} \sum_{i=0}^{k-1} \left(\frac{1}{2}\right)^i \ln(\sqrt{2}) = \left[\sum_{i=0}^{\infty} \left(\frac{1}{2}\right)^i\right] \ln(\sqrt{2}) = 2 \ln (\sqrt{2}) = \ln (2)$$

and, finally, we conclude that $x_{\infty} = \exp(\tilde{x}_{\infty}) = 2$.

Let us denote your sequence by $x_n$. Then $$x_1=\sqrt{2},\ x_{n+1}=\sqrt{2x_n} \quad \forall\ n \ge 1.$$ Clearly $\sqrt{2}\le x_n<2$ for every $n \ge 1$, and $$x_n-x_{n+1}=x_n-\sqrt{2x_n}=\frac{x_n^2-2x_n}{x_n+x_{n+1}}=\frac{x_n(x_n-2)}{x_n+x_{n+1}}<0 \quad \forall n \ge 1.$$ Hence $(x_n)$ is increasing and bounded above. It follows that $(x_n)$ is convergent. If $l$ denotes its limit, then $\sqrt{2}\le l \le 2$ and $l=\sqrt{2l}$. Solving the equation $l=\sqrt{2l}$ we get $l\in \{0,2\}$, and since $\sqrt{2} \le l \le 2$, we deduce that $l=2$.

You can also observe that

$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} \cdot \sqrt{\sqrt{\sqrt{...\sqrt{2}}}}=2$$

thus

$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} =\frac{2}{\sqrt[2^n]{2}}$$
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