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Find the value of $$\sqrt{1\sqrt{2\sqrt {3 \sqrt{4\sqrt{5\sqrt{6\cdots\sqrt{\infty}}}}}}}$$

What is the absolute value of the root in below question and what does it represent geometrically, I had a few approaches leading to possible values to approximation and know the answers, but I also need what it represents geometrically.

Did
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chilchin
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    Please use more informative titles. And also don't misuse tags. This question has absolutely nothing to do with either (functional-analysis) or (complex-analysis). – Willie Wong Apr 10 '13 at 10:53
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    this forum has a collaborative editing feature. If you don't want people to edit your posts, don't post here. – Willie Wong Apr 10 '13 at 10:59
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    Well somehow, chilchin isn't only rude towards me. I'm relieved to hear that... – Thomas Apr 10 '13 at 11:05
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    I removed many comments here too that were not at all constructive. chilchin, I see that you are very aggressive and rude in comments. This is simply not appropriate here. – davidlowryduda Apr 10 '13 at 11:13
  • @chilchin: what were your approaches to this problem? There doesn't seem to be much room for maneuver to me. – Ron Gordon Apr 10 '13 at 11:13
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    This really is an interesting question, but the way it is presented just makes it unattractive. – Thomas Apr 10 '13 at 11:15
  • This (https://oeis.org/A112302), its square (https://oeis.org/A259235), and its logarithm (https://oeis.org/A114124) can be found on OEIS. – Akiva Weinberger May 17 '24 at 04:09

3 Answers3

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Let the limiting value of the above expression be $L$; then

$$\log{L} = \sum_{k=1}^{\infty} \frac{\log{k}}{2^k}$$

Ron Gordon
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  • With $\log L \approx 0.507833922868439$. I had the same idea. WolframAlpha states that this series does converge. – Thomas Apr 10 '13 at 11:07
  • yeah in my approaches and it approaches i get values and some of us have researched it but i think there is fundamental flaw there and i also believe infinite vlaues always approach to a finite value:) – chilchin Apr 10 '13 at 11:08
  • @macydanim: it should be clear that the expression converges by comparison with $$\sum_{k=1}^{\infty} \frac{k^{\epsilon}}{2^k}$$ – Ron Gordon Apr 10 '13 at 11:09
  • @RonGordon Thanks for the information. You are right it should be clear. – Thomas Apr 10 '13 at 11:10
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    @Ron: if the above expression is interpreted as $$\lim_{n \to \infty} \sqrt{1\sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}}}$$ then it does in fact have a limiting value as you observed. (This is to say, under what I think of as the natural interpretation, you don't need the word "If" in your answer.) – Willie Wong Apr 10 '13 at 11:20
  • ALL MY COMMENTS ARE DELETED HERE BY SOME ADMIN WHAT CAN I DO? @WILLIE no thew limit u suggested is not my question my question is aBSOLUTE – chilchin Apr 10 '13 at 11:24
  • @willie: I agree, but I was hedging my bets and was not prepared to proved that the limit exists. – Ron Gordon Apr 10 '13 at 12:08
  • @chilchin: what do you mean by "absolute value" that I am not getting? It should be clear that the posted expression is positive. If you mean the explicit value, then anyone on this site should see that one obtains the value of $L$ by exponentiating the above result. – Ron Gordon Apr 10 '13 at 12:43
  • i didnot mean in that sense, absolute here being final and resolute my question posted here is only a part of what is an analytically rational and complex function of more than 1 variable i thought if ppl could answer as well as some students and experts did on some other site i could have provided u ,should i say its sibling function and they magically in one approach tend to get already know very very famous ratio or constant ..but anyhow dont ignore this one one of the values suggested by what we got also tend to have characteristics of a well known crop circle njoy ur time byeee :))) – chilchin Apr 10 '13 at 12:57
  • @StevenGregory: The alleged duplicate was asked a year after this one. Think again. – Ron Gordon May 01 '15 at 21:39
  • OK. I thought again. See this for part of the answer. – Steven Alexis Gregory May 01 '15 at 22:19
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As indicated by Ron Gordon the logarithm of this is : $$l= \sum_{k=1}^\infty \frac{\log{k}}{2^k}$$ Consider the polylogarithm function : $$f(s):=\sum_{k=1}^\infty \frac{(1/2)^k}{k^s}=\operatorname{Li}_s\left(\frac 12\right)$$ then $$f'(s)=-\sum_{k=1}^\infty \frac{\log(k)}{2^k\;k^s}$$ so that the final answer may be written as $\ \boxed{\displaystyle e^{-\operatorname{Li}_{0'}\left(1/2\right)}}$ with the meaning : $e^{-\lim_{s\to 0^+}\frac d{ds}\operatorname{Li}_s\left(1/2\right)}$

Raymond Manzoni
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  • Nice(+1). I was just thinking whether I could interpret it as the derivative of some dirichlet-like series, but couldn't find one quickly. So this is an additional result. Cool... – Gottfried Helms May 20 '13 at 21:52
  • Thanks @Gottfried ! You could write it too as derivative of the Lerch zeta function. Unfortunately pari seems unable to evaluate non integer first terms (Wolfram Alpha is clearly better and returns the correct numerical value). – Raymond Manzoni May 20 '13 at 22:32
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The number in question is called Somos's Quadratic Recurrence Constant, whose approximate value can be found here. It is similar to the Nested Radical Constant.

Lucian
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