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Is there any related studies about the topic?

Jyrki Lahtonen
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Cleo Tejano
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    Related (if not duplicate): https://math.stackexchange.com/q/1659502, https://math.stackexchange.com/q/1785454. – Martin R Dec 04 '19 at 13:33

1 Answers1

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Let $$n=\sqrt{k+\sqrt{k^2+\sqrt{k^3+\cdots}}}$$ then $$n^2=k+\sqrt{k^2+\sqrt{k^3+\cdots}}$$ By subtracting $k$ from both sides and then multiplying out $\sqrt{k}$ we get $$n^2 - k=\sqrt{k}\sqrt{k+\sqrt{k^2+\sqrt{k^3+\cdots}}}$$ But now we got the same form as in the beginning, thus we can insert $n$ to get $$n^2 - k=\sqrt{k}\cdot n$$ with the solutions $$n_{1,2}=\frac{\sqrt{k}}{2} \pm \sqrt{\frac{k^2}{2}+k}$$

Its strange that there are two solutions, so please check my solution for errors. But the main idea in these kind of questions is always to get back to the first form.

Nurator
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  • @TodorMarkov Are you sure? Can you elaborate why it should work for this but not with one more term? – Nurator Dec 04 '19 at 13:43
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    When you go from equation 2 to equation 3, your $k^2$ does become $k$, but your $k^3$ should also become $k$, while the $k^4$ in the $\dots$ should become $1$. – Todor Markov Dec 04 '19 at 13:43
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    You only need two terms to see that this answer is wrong: $\sqrt k\sqrt{k+\sqrt{k^2}}$ is not the same as $\sqrt{k^2+\sqrt{k^3}}$, because $k\sqrt{k^2}\ne\sqrt{k^3}$. – TonyK Dec 04 '19 at 13:44
  • Yes you are both right. Thank you for pointing that out! Any idea how to correct the proof? – Nurator Dec 04 '19 at 13:46
  • Choose + sign in the final step so that $n$ is positive definite for positive values of $k$. – Z Ahmed Dec 04 '19 at 14:22
  • This is an interesting, but unfixably flawed attempt. – David Diaz Dec 04 '19 at 15:48
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    @DavidDiaz That is the most polite "You are wrong" I have ever read :D I will use that quote in my every day life now! – Nurator Dec 04 '19 at 15:53