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i'd like to know how to prove that following sequence

$$\sqrt{2\sqrt2},\sqrt{2\sqrt{2\sqrt2}},\sqrt{2\sqrt{2\sqrt{2\sqrt2}}},...$$

Is convergent, and that it's limit equals two.

In other words, that $$\sqrt{2\sqrt{2\sqrt{2\sqrt2...}}}=2$$

I imagine it's supposed to be simple and straightfoward, but i cant figure it out. I would be grateful for any help you could give me.

MJD
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B. de Morais
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3 Answers3

8

Your sequence could be rewritten as:

$$ \begin{cases} a_0=\sqrt{2}\\ a_{n+1}=\sqrt{2a_n} \end{cases} $$

By induction, you can easily prove that:

$$ a_n=2^{\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}}=2^{(1-\frac{1}{2^n})}, n\geq 1 $$

The sum in the power converges, hence the sequence converges and the limit is $2$

Oussama Boussif
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  • Thanks for answering! I did something similar the first time i tried this, the problem is that i find

    $ a_n=2^{\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}}=2^{\sum{(\frac{1}{2^n})}} $

    Where i know that the series $\sum{(\frac{1}{2^n})}$ converges to 2. How do i find that

    ${\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}}={(1-\frac{1}{2^n})}, n\geq 1 $? What am i missing?

     – B. de Morais May 27 '17 at 22:20
  • @BrM Careful: where you know the series converges to $;1;$ ...observe that geometric series begins with $;n=1;$ ... – DonAntonio May 27 '17 at 22:22
  • ooooooh..... * facepalm * – B. de Morais May 27 '17 at 22:32
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Here's a non-rigorous way to find the answer. Call the final result $x$. Then $x^2 = 2 \cdot x$, and you can solve for $x$. This method can give the wrong answer for similar-looking problems: the iteration can converge to different values depending on the starting point, or it may not converge at all.

Dan Brumleve
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    It's worth pointing out that $0$ is also a solution to this equation, and so when checking the sequence converges, you'd also need to check it converges to the right root. – B. Mehta May 27 '17 at 21:47
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    B. Mehta indeed, and without specifying the meaning of the expression $\sqrt{2 \cdot \sqrt{2 \cdot \sqrt{2 \cdot \dots}}}$ as a limit of certain finite expressions, it's arguably equally valid to say that it equals $0$. – Dan Brumleve May 27 '17 at 21:51
  • This answer and Don Antonio 's answer form a complete solution – Matthew Leingang May 28 '17 at 01:38
  • I think the method is rigorous, right? You just have to remember it's a one-way implication... or is it actually not rigorous? – user541686 May 28 '17 at 04:55
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An idea: define

$$x_1:=\sqrt2\;,\;\;x_2=\sqrt{2x_1}\,,\ldots,x_n:=\sqrt{2x_{n-1}}$$

Show that $\;\{x_n\}\;$ is (1) bounded and (2) monotone ascending, and thus it has a limit. Now use arithmetic of limits.

B. Mehta
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DonAntonio
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