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$\begingroup$

This is a spinoff of this question

Defining

$$f_0(x) = x$$ $$f_n(x) = \log(f_{(n-1)} (x)) \space (\forall n>0)$$

and

$$a_0 = 1$$ $$a_{n+1} = (n+1)^{a_n} \space (\forall n>0)$$

How to calculate

$$\lim_{n \to \infty } f_n(a_n) $$

(an "experiment" here, but (beware) I think WolframAlpha is using an approximate representation in powers of 10)

Edit

A table with the first few values (made with aid of hypercalc, as per Gottfried's suggestion in comments)

$$\begin{array}{cc} n & f_n(a_n) \\ 0 & 1. \\ 1 & 0. \\ 2 & -0.366513 \\ 3 & -0.239279 \\ 4 & -0.0771089 \\ 5 & -0.06133128660211943 \\ 6 & -0.06133124230008346 \\ 7 & -0.06133124230008346 \\ 8 & -0.06133124230008346 \end{array}$$

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    (prev. comment removed) It does look like it converges! $f_5$,$f_6$, and $f_7$ all are approximately -.0613312.2012-10-20
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    Haven't been able to prove it yet. It's definitely monotonically increasing: For all $n>2$ we have $n>e$, so that $n^{a_{n-1}}=a_n>e^{a_{n-1}}$. Then $f_n(a_n)>f_n(e^{a_{n-1}})=f_{n-1}(a_{n-1})$. I'm stuck on proving that it's bounded. You can see that sums of the form $e-2\log(3)$, $e^e-3^2\log(4)$, $e^{e^e}-4^{3^2}\log(5)$, etc. would all have to be greater than zero for boundedness, but I'm having trouble proving it. Clearly the condition depends strongly on the magnitude of $e$ though.2012-10-20
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    by "having trouble proving it" I meant "Having trouble proving that all those terms are greater than zero"2012-10-20
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    @belisarius: I'm not sure about your numerical values. I get the following 1, 0, -0.3665129206, -0.2392792745, 0.04639873165, 0.2116219343, 0.2785508275, 0.3021805399, 0.3099107249, 0.3123003883, etc.2012-10-21
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    @SidiousLord Yep. I got some index displacement in my table. Corrected, thanks2012-10-21
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    Have you tried the "hypercalc" at Robert Manufo's site? Possibly one can get more terms of your sequence with that...2012-10-22
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    @GottfriedHelms Nice suggestion. Done! Seems it converges fast!2012-10-22
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    An interesting variant seems to be the sequence $\log(2) $ , $\log(\log(2^3)) $ , $ \log(\log(\log(2^{3^4}))) $ , ... ($\log $ means here the *ln* ). Does it also converge? I went up to top exponent 16 on the hypercalc-site, but didn't arrive at a conclusion....2012-10-22
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    I went upt to top exponent $(2^7) $ on the hypercalc site stepping to $2^k$ as the top-exponents. With the function $g(x) = a= 2\^3\^4\^...\^(2^x); b=\ln°^{2^x-1}(a) ; return(b) $ (where the superscript at the ln indicates iterations) I get the following table: $\begin{array} {} g(1) = 0.6931471805599453 \\ g(2) = 1.393254144068342 \\ g(3) = 2.781949728441699 \\ g(4) = 3.7909671825168587 \\ g(5) = 4.710536164389033 \\ g(6) = 5.58566765572222 \\ g(7) = 6.432348427854444 \\ \end{array} \\ $2012-10-22
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    ... and two more : $\begin{array} {} g(8) = 7.258605780274824 \\ g(9) = 8.069295828717253 \\ \end{array} \\ $ . Hmmm...2012-10-22

4 Answers 4