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Let $a_0=1$ and $a_n=1+\sqrt{a_{n-1}}+\sqrt{1+\sqrt{a_{n-1}}}$

I want to know if the limit of $a_n$ as n goes to infinity. $$1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}}}$$ When I tried to find a possible answer I tried to solve $1+\sqrt{x}+\sqrt{1+\sqrt{x}}=x$. I tried for an hour so I gave up and asked WolframAlpha and it got $x≈5.04891733952231$

I'm not sure how to find out if $a_n$ converges. I think it does

  • Your "infinite radical" doesn't have any meaning all on its own, so it's not clear what you are asking. If it could be rewritten as a sequence of actual real numbers then perhaps this question would make sense. There are a bunch of similar posts where this is done properly, for example https://math.stackexchange.com/questions/1109537/sums-of-nested-radicals?rq=1, https://math.stackexchange.com/questions/61012/nested-square-roots-50-sqrt51-sqrt52-sqrt54-sqrt-dots?rq=1, and others. – Lee Mosher Jun 07 '21 at 14:29
  • @LeeMosher I think I can rewrite this problem –  Jun 07 '21 at 14:38
  • If you substitute $y=\sqrt x$, you might be able to find a cubic that $y$ satisfies. – Empy2 Jun 07 '21 at 14:40
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    @Empy2 okay let me try to do that in this comment $1+y+\sqrt{1+y}=y^2$ then I could move all the none radicals to one side $1+y-y^2=\sqrt{1+y}$ then I square it all. $y^4 - 2 y^3 - y^2 + 2 y + 1=1+y$ which is quantic but then if I put it all on one side I get $y(y^3-2y^2-y+1)=0$ I still get a quantic expression where do I go from here? –  Jun 07 '21 at 14:52
  • @Empy2 never mind because $y$ isn't zero we know that $y^3-2y-y+1=0$ so I guess you are right. –  Jun 07 '21 at 14:57
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    For your own knowledge, this sequence converges to $2+2 \sqrt{\frac{7}{3}} \cos \left(\frac{1}{3} \tan ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\right)$. Mathematica was able to give a full solution to $0=1+\sqrt{x}+\sqrt{1+\sqrt{x}}-x$ – QC_QAOA Jun 07 '21 at 15:20
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    FWIW, I got the answer $$\frac{4 \left(1 + \sqrt{7} \cos{\left(\frac{\arctan{\left(3 \sqrt{3} \right)}}{3} \right)}\right)^{2}}{9}$$ using sympy – saulspatz Jun 07 '21 at 15:30
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    @saulspatz it is also $ 1 + 4 \cos \frac{2 \pi}{7} + 4 \cos^2 \frac{2 \pi}{7} = \left( 1 + 2 \cos \frac{2 \pi}{7} \right)^2 ; ; ; ; $ – Will Jagy Jun 07 '21 at 16:00
  • @WillJagy Much nicer looking. How did you find it? – saulspatz Jun 07 '21 at 16:12
  • @saulspatz I wrote $x = w^2$ and reached $w^3 - 2 w^2 - w + 1 = 0$ I noticed this was Casus Irreducibilis; then I got suspicious and asked about the Galois group of the cubic, which is cyclic, also discriminant $49$ is square. Next $w=t+1$ gave $t^3 + t^2 - 2t + 1,$ and I remembered the roots of that (twice some cosines), straightforward with Gausses methods for cyclotomic things, also can be done with trigonometry identities see http://zakuski.utsa.edu/~jagy/cox_galois_Gaussian_periods.pdf – Will Jagy Jun 07 '21 at 16:18
  • @WillJagy Impressive. – saulspatz Jun 07 '21 at 16:31
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    @saulspatz thanks. Here is a question on cubics where I put a big list http://math.stackexchange.com/questions/2022216/on-the-trigonometric-roots-of-a-cubic/2022887#2022887 , Tito asked an early one here https://math.stackexchange.com/questions/1065862/something-strange-about-sqrt-4-sqrt-4-sqrt-4-x-x-and-its-friends Numerous examples in Reuschle https://www.google.com/books/edition/Tafeln_complexer_Primzahlen/wt7lgfeYqMQC?hl=en&gbpv=1&dq=reuschle++tafeln+complexer+primzahlen&pg=PR1&printsec=frontcover#v=onepage&q=reuschle%20%20tafeln%20complexer%20primzahlen&f=false – Will Jagy Jun 07 '21 at 16:41

2 Answers2

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Notice the map $x\mapsto f(x) \stackrel{def}{=} 1 + \sqrt{x} + \sqrt{1 + \sqrt{x}}$ is defined and strictly increasing for $x \ge 0$.

Since $a_1 > a_0$, we have $$a_2 = f(a_1) > f(a_0) = a_1 \implies a_3 = f(a_2) > f(a_1) = a_2 \implies \cdots $$ So $a_n$ is an increasing sequence.

The map $f(x)$ has a fixed point at $b \sim 5.05$ (in fact, this is the first non-zero fixed point of $f(x)$). Since $a_0 < b$, we have $$a_1 = f(a_0) < f(b) = b \implies a_2 = f(a_1) < f(b) = b \implies \cdots$$ So $a_n$ is also bounded from above by $b$. As a result, $a_n$ converges to some value $a_\infty$ in $(a_1,b] = (1,b]$.

Now $f(x)$ is a continuous function, this implies

$$f(a_\infty) = f(\lim_{n\to\infty}a_n) = \lim_{n\to\infty} f(a_n) = \lim_{n\to\infty} a_{n+1} = a_\infty$$

So $a_\infty$ is also a fixed point of $f(x)$ too. Since $b$ is the first non-zero fixed point of $f(x)$, this forces $a_\infty = b$. ie. your sequence $a_n$ converges to the fixed point you have.

achille hui
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Starting from @Will Jagy first hint $$w^3 - 2 w^2 - w + 1 = 0\qquad \text{with} \qquad x=w^2$$ the cubic shows three real roots $(\Delta=49)$. Using the trigonometric method, they are $$w_k=\frac{2}{3} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \left(2 \pi k-\sec ^{-1}\left(2 \sqrt{7}\right)\right)\right)\right)\qquad \text{with} \qquad k=0,1,2$$ and the solution of concern is $w_0$ (not prooved).

This gives $$x=\frac{4}{9} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \sec ^{-1}\left(2 \sqrt{7}\right)\right)\right)^2$$ which is another expression of the result.

Claude Leibovici
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  • Other roots of the cubic will give the answer with different choices for the signs of the square-roots – Empy2 Jun 08 '21 at 15:05