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


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.


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 $2$

  • $\begingroup$ 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? $\endgroup$ – B. de Morais May 27 '17 at 22:20
  • $\begingroup$ @BrM Careful: where you know the series converges to $\;1\;$ ...observe that geometric series begins with $\;n=1\;$ ... $\endgroup$ – DonAntonio May 27 '17 at 22:22
  • $\begingroup$ ooooooh..... * facepalm * $\endgroup$ – B. de Morais May 27 '17 at 22:32

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.

  • 6
    $\begingroup$ 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. $\endgroup$ – B. Mehta May 27 '17 at 21:47
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    $\begingroup$ 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 $0$. $\endgroup$ – Dan Brumleve May 27 '17 at 21:51
  • $\begingroup$ This answer and Don Antonio 's answer form a complete solution $\endgroup$ – Matthew Leingang May 28 '17 at 1:38
  • $\begingroup$ I think the method is rigorous, right? You just have to remember it's a one-way implication... or is it actually not rigorous? $\endgroup$ – Mehrdad May 28 '17 at 4:55

An idea: define


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


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