1

$$\sqrt{1+\sqrt{2+\sqrt{3+\ldots}}}$$

My approach

1st approach: Let $\sqrt{1+\sqrt{2+a}} = x$, where $a$ is a result of the lower portion of the continued nested radical and $1+\sqrt{2+a}$ is a perfect square, and $x$ is a number that doesn't contain a radical.

Solving for $a$ gives 7, and $x$ equals to 2.

2nd approach: Using a software, $\sqrt{1+\sqrt{2+\sqrt{3+\ldots}}} \approx 1.75793...$

What is the best approach for the nested radicals?

MMJM
  • 637
  • I think it is enough for me to say $1.757... \approx 2$ – MMJM Jan 19 '19 at 05:05
  • Marking this as duplicate – MMJM Jan 19 '19 at 05:06
  • How on earth did you solve one equation with two variables? –  Jan 19 '19 at 07:50
  • neglect x, since x is the result of $\sqrt{1+\sqrt{2+a}}$, where $1 + \sqrt{2+a}$ is a perfect square. $x$ just represent the result of the equation here – MMJM Jan 19 '19 at 08:03
  • I do not yet understand.. can you write that as edit to your question... Are you saying $1+\sqrt{2+a}$ is a perfect square only for one value of $a$?? How do you even known that $a=\sqrt{3+\sqrt{4+\cdots}}$ even converge? –  Jan 19 '19 at 08:04
  • @Praphulla I'll just give $x$ a description for it. It does not mean that I'll solve for both variables. The $a$ is the only value to be calculated. – MMJM Jan 19 '19 at 08:08
  • @Praphulla for your 1st question: No, I give it a small number(not fraction or decimal) that satisfy $\sqrt{1+\sqrt{2 + a}}$. I mean to cancel all the radicals and shoot the values to a single number. – MMJM Jan 19 '19 at 08:21
  • I think you are missing the point... All I am asking is how do you know $\sqrt{3+\sqrt{4+\sqrt{5+\cdots}}}$ is a real number? How do you know it converge?? –  Jan 19 '19 at 08:38
  • I dont know it. I just assumed $\sqrt{3+\sqrt{4+\sqrt{5+\cdots}}}$ is a real number and converge. – MMJM Jan 20 '19 at 01:56
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    You are not supposed to assume something which you do not know that it exists.,, –  Jan 20 '19 at 03:10

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