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$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$$ I was working on this series from the UKMT website. I solved it to get the two roots. (1 + sqr5)/2 (Golden ratio number).

Now I am asking myself what assumptions did I make in this calculation? Does the series converge, absolutely and what is the region then of convergence?

Nosrati
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Waitara Mburu
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  • At the very least, you're assuming that the limit exists. – Darius May 12 '19 at 23:09
  • If you assume that the nth term is golden ratio to some power, can you inductively prove that this is correct? Then you can prove the limit exists – fGDu94 May 12 '19 at 23:57
  • If you consider it as the limit of the sequence $$\sqrt{1}, \sqrt{1+\sqrt{1}}, \sqrt{1+\sqrt{1+\sqrt{1}}},\dots$$ then since the sequence is clearly increasing it's enough to show that it is bounded above. – saulspatz May 13 '19 at 00:03
  • What series are you talking about? Do you have a power series secluded behind the curtain? – Ted Shifrin May 13 '19 at 00:52
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    "region of convergence" doesn't make sense for an expression without a variable in it somewhere. – Gerry Myerson May 13 '19 at 01:16
  • Somewhat related: https://math.stackexchange.com/questions/2255364/crazy-iterated-square-roots and https://math.stackexchange.com/questions/888130/equality-of-two-iterated-square-roots – Gerry Myerson May 13 '19 at 01:18
  • Also https://math.stackexchange.com/questions/2132412/is-there-an-explicit-formula-that-gives-the-value-of-sqrt2-sqrt2-sqrt2-c/2132529 and https://math.stackexchange.com/questions/1005434/possible-values-of-infinitely-nested-square-root-n-sqrtx-sqrtx-sqrtx-sq and https://math.stackexchange.com/questions/654478/calculate-limit-0f-nested-square-roots – Gerry Myerson May 13 '19 at 01:23
  • Please try to engage with the comments, Waitara. – Gerry Myerson May 14 '19 at 03:18
  • Are you still here, Waitara? – Gerry Myerson May 15 '19 at 05:54
  • Sorry my sabaticle! very helpful indeed Gerry! Links very helpful,still on it, looks like we need 1 in the nested roots, otherwise convergence not guaranteed! – Waitara Mburu May 16 '19 at 01:18

1 Answers1

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Let $a$ be our sequence.

Thus, $a_1=1,$ $a_{n+1}=\sqrt{1+a_n}$, $a_n\geq1$ and by your work we obtain: $$\left|a_n-\frac{1+\sqrt5}{2}\right|=\left|\sqrt{1+a_{n-1}}-\frac{1+\sqrt5}{2}\right|=\frac{\left|1+a_{n-1}-\frac{3+\sqrt5}{2}\right|}{\sqrt{1+a_{n-1}}+\frac{1+\sqrt5}{2}}=$$ $$=\frac{\left|a_{n-1}-\frac{1+\sqrt5}{2}\right|}{\sqrt{1+a_{n-1}}+\frac{1+\sqrt5}{2}}\leq\frac{\left|a_{n-1}-\frac{1+\sqrt5}{2}\right|}{\sqrt{2}+\frac{1+\sqrt5}{2}}\leq\frac{1}{3}\left|a_{n-1}-\frac{1+\sqrt5}{2}\right|\leq$$ $$\leq\frac{1}{3^2}\left|a_{n-2}-\frac{1+\sqrt5}{2}\right|\leq...\leq\frac{1}{3^{n-1}}\left|a_1-\frac{1+\sqrt5}{2}\right|\rightarrow0,$$ which says that indeed $$\lim_{n\rightarrow+\infty}a_n=\frac{1+\sqrt5}{2}.$$

Michael Rozenberg
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  • so if we call the limit, some value, K, then a general relation K = (1 + K) ^1/2 must be satisfied. Next, we could use inductive methods as George suggests above right? – Waitara Mburu May 13 '19 at 22:18
  • @Waitara Mburu It works for your problem, but we need to release this way. I showed another idea. – Michael Rozenberg May 14 '19 at 02:39