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Evaluate the following expression to 4 decimal places:

$$\sqrt{1+2\sqrt{1+2\sqrt{1+\dots+2\sqrt{1+2\sqrt{2015}}}}}$$

where the number of square roots in the expression is exactly 2015.

As per the tag, you may not use a computer, calculator, or any electronic aid in your calculations. You may look up the decimal expansion of a certain constant, but the people answering this question will probably have it memorised to several decimal places anyway!

I believe I found this puzzle somewhere on the internet, rather than having made it up myself, but after such a long time I can't remember where. If anyone finds this puzzle elsewhere on the web, please drop a link in a comment and I'll edit it in.

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  • $\begingroup$ How many iterations are in the ...? $\endgroup$ – xnor Apr 13 '15 at 0:30
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    $\begingroup$ @xnor Enough to make the number of square roots 2015. $\endgroup$ – LeppyR64 Apr 13 '15 at 0:44
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    $\begingroup$ @JasonLepack 2010 :c) $\endgroup$ – BmyGuest Apr 13 '15 at 5:36
  • $\begingroup$ A "no-computers" problem would be more elegant if you didn't have to look up any constants -- I'm not going to downvote the question or anything but it would have been nicer to rig it so the constant that came out was an integer. $\endgroup$ – Brian May 28 '15 at 3:28
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2.4142

Proof

To restate the problem, we are given

$a_1=\sqrt{2015}$

$a_{n+1} = \sqrt{1+2a_n}$

and asked to find $a_{2015}$.

Suppose this sequence has a limit $L = \lim_{n\rightarrow \infty} a_n$; then

$L = \sqrt{1+2L}$

which has solution

$L = 1+\sqrt{2}$.

Now look at the difference between $a_n$ and $L$. If $a_n = L+\Delta$, then a Newton's approximation gives

$a_{n+1} = \sqrt{1+2+2\sqrt{2}+2\Delta} < 1+\sqrt{2} + \frac{\Delta}{1+\sqrt{2}}$.

So by induction we can show that for all $n$,

$L < a_n \leq L + (\sqrt{2015}-1-\sqrt{2}) \left(1+\sqrt{2}\right)^{1-n}$.

In particular,

$1+\sqrt{2} < a_{2015} < 1+\sqrt{2}+ (\sqrt{2015} -1-\sqrt{2})\left(1+\sqrt{2}\right)^{-2014}$.

$(\sqrt{2015}-1-\sqrt{2}) \left(1+\sqrt{2}\right)^{-2014}$ is much less than 0.0001, so it can be ignored when stating the answer to the original question.

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  • $\begingroup$ Excellent answer and proof! +1 and accept. $\endgroup$ – Rand al'Thor Apr 13 '15 at 10:05
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The expression is an approximation of the equation $1 + 2x = x^2$ or $x^2 - 2x - 1 = 0$, so I'm assuming that the answer to it is $x$, which is $\frac{2 + \sqrt{8}}{2}$ or $1 + \sqrt{2}$. This is equal to approximately 2.4142.

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  • $\begingroup$ Well done on being first to the right answer! But you didn't give a full proof, so I'll have to accept aschepler's answer instead. $\endgroup$ – Rand al'Thor Apr 13 '15 at 9:42
  • $\begingroup$ That's alright. $\endgroup$ – Joe Z. Apr 13 '15 at 16:35

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