Evaluate the nested square roots

Question

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 no-computers 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.}

Answer 1

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.

Answer 2

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.