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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. –  NeuroFuzzy Oct 20 '12 at 4:20
    
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. –  NeuroFuzzy Oct 20 '12 at 5:22
    
by "having trouble proving it" I meant "Having trouble proving that all those terms are greater than zero" –  NeuroFuzzy Oct 20 '12 at 5:29
    
@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. –  Sidious Lord Oct 21 '12 at 9:40
    
@SidiousLord Yep. I got some index displacement in my table. Corrected, thanks –  belisarius Oct 21 '12 at 13:52
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4 Answers

This is not an answer as one would like to write one; but there is still only one other answer around so far - thus a second attempt might be helpful anyway even if not yet complete.

I've come across a claim, that Ramanujan has proved, that the function $$ f(x) = 1 + {e^x \over 2^3} + {e^{e^x} \over 2^{3^4}} + {e^{e^{e^x}} \over 2^{3^{4^5}}} + ... $$ is entire. This implies, that the sequence of terms approach zero for each x, and thus that the denominators outgrow the numerators. By this we see, that the iterated logarithms applied to the numerators, when iterated h times until they arrive at x, and then applied with the same number of iterations to the denominators must arrive at some value arbitrarily greater than x for each k'th term after $ k \gt K$ for some fixed $K $.

Thus we have the first information, that $$ \ln^{[h]} \left(2^{3^{4^{\ldots ^h}}} \right)$$ where $ \ln^{[h]} $ means the h'fold iteration of the $\ln $ , diverges with increasing h .

The second information is the trivial one, that if the entries in the iterated exponential are all equal, and specifically equal the base of natural logarithms $e$ , $$ \ln^{[h]} \left(e^{e^{e^{\ldots ^e}}} \right)$$ then the iterated logarithm of the iterated exponential always arrives at the same constant value for increasing h.

The third information is then, that the iterated exponential in the initial question has decreasing terms, $$ \ln^{[h]} \left(h^{\ldots ^{4^{3^2}}} \right)$$ so possibly using the argumentation of Ramanujan one can also prove, that the expression asked for, actually converges to a fixed value. Unfortunately I don't have access to that argumentation of Ramanujan, but here is a link in the tetration-forum to (possibly) more helpful information how to retrieve that source.

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Your sequence is convergent :

Fist, it is well-defined, because it is defined for $n=2$, and for $n\ge2$, $\log^{n+1}(a_{n+1}) = \log^n(a_n \log(n+1)) = \log^{n-1}(\log a_n + \log^2(n+1))$ and $\log^2(n+1) > 0$, which also shows that the sequence is increasing from that point.

Let $x_n = \log^n(a_n)$. By using that for all $x,y > 0, \log(x+y) \le \log x + y/x$ and that $\log$ is increasing, we get :

$x_{n+1} = \log^{n-1}(\log a_n + \log^2(n+1)) \le \log^{n-2}(\log^2 a_n + \log^2(n+1)/ \log a_n) \\ \le \ldots \le \log^n(a_n) + \log^2(n+1)/ (\log a_n \log^2 a_n \ldots \log^{n-1} a_n) \\ = x_n + \log^2(n+1)/(\exp(x_n) \ldots \exp^{n-1}(x_n))$

And since we know $x_n$ is increasing, it is easy to show that $x_{n+1} - x_n$ is summable, and that it converges extremely fast (the number of correct digits grows tetrationnally). There should not be enough room in the observable universe to write all the correct digits of $x_7$

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You indeed have x_n = log^(n) ( n^...^1 )

= log^(n-2) ( (n-2)^...^1 * log(n-1) + log( log( n )))

which does not simplify further, but allows asymptotical expansion, via

log( A+B ) = log( A ) + log(1 + B/A) = log(A) - sum((-B/A)^k, k=1..oo).

Unfortunatley, this is inside the "innermost" of the n-2 exterior logs. However, the log(log n) term is quickly negligable w.r.t. the (n-2)^...^1 log(n-1). Therefore, log^(n-2) ( (n-2)^...^1 * log(n-1)) is a very good approximation to x_n, for large n. For the same reason, neglecting the log(n-1) yields still a very good approximation to x_n, for large n. But this means we end up with x_(n-2) as very good approximation to x_n. This can be seen as "heuristic" (well, maybe just a little better than "hand waving") justification of the convergence.

OTOH, quantification of the "very good" is possible, and yields a proof. (Probably the same as above.)

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I don't think this sequence converges. First note that $f_n(a_n)=f_{n-1}(a_{n-1})+log(...(log(n))...)$ where log is composed n-times. Then note that

$\lim_{n\to\infty}log(...(log(n))...)=0$ where log is composed n-times. Then if $f_n(a_n)$ converges then $-f_n(a_n)$ converges as well. With that notice

$\lim_{n\to\infty}-f_n(a_n)=\sum_{k=n}^{\infty}(f_{k+1}(a_{k+1})-f_k(a_k))=\lim_{n\to\infty}\sum_{k=n}^{\infty}\int_{1}^{log(...(log(k))...))}\frac{1}{x}dx =\lim_{n\to\infty}-\sum_{k=n}^{\infty}\int_{log(...(log(k))...))}^{1}\frac{1}{x}dx=\int_{0}^{1}\frac{1}{x}$

where log is composed k-times in the integral.

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Log[Log[Log[3^2]]] evaluates to -0.239279..., Log[Log[2]] + Log[Log[Log[3]]] evaluates to -2.73046 (same if you just add Log[Log[3]], composing n-1 times). We have $\log(\log(\log(3^2)))=\log(\log(2\log(3)))=\log(\log(2)+\log(\log(3)))$, yes, but you can't move that addition outside that last logarithm. –  NeuroFuzzy Oct 20 '12 at 6:33
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