My Math Forum Limit involving 1/2-height superexponential

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 April 28th, 2013, 02:04 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Limit involving 1/2-height superexponential Hi, Does the limit converges : $\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \sum_{i=1}^{j} \frac{1}{{}^{1/2} i}$ I initially guessed that it's somewhat << log(n) but I think it can be tighten up to << n^? for some reasonably small ? > 0. Thanks, Balarka .
 April 28th, 2013, 07:21 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Limit involving 1/2-height superexponential What definition are you using for $^yx$? I'm accustomed to seeing only the restriction to the naturals for the superexponent (though I've seen a few attempts to generalize it). Unrelated: this is the single ugliest notation I know of in mathematics.
April 28th, 2013, 07:49 AM   #3
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Re: Limit involving 1/2-height superexponential

Quote:
 Originally Posted by CRGreathouse What definition are you using for $^yx$?
Kneser's analytic continuation.

Quote:
 Originally Posted by CRGreathouse Unrelated: this is the single ugliest notation I know of in mathematics.
I thought it was the most familiar notation to you . . . Okay, from now on, I will use sexp(x, y).

April 29th, 2013, 05:56 AM   #4
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Re: Limit involving 1/2-height superexponential

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by CRGreathouse Unrelated: this is the single ugliest notation I know of in mathematics.
I thought it was the most familiar notation to you . . . Okay, from now on, I will use sexp(x, y).
Familiar, yes, but not very nice. It doesn't scan easily with its similarity to the usual notation for exponentiation.

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by CRGreathouse What definition are you using for $^yx$?
Kneser's analytic continuation.
This one?
http://www.digizeitschriften.de/dms/img ... N002175851

April 29th, 2013, 06:45 AM   #5
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Re: Limit involving 1/2-height superexponential

Quote:
 Originally Posted by CRGreathouse It doesn't scan easily with its similarity to the usual notation for exponentiation.
Yes, and it also is similar to y with a subscript x, no?

Quote:
 Originally Posted by CRGreathouse This one? http://www.digizeitschriften.de/dms/img ... N002175851
Yes, exactly that one.

Quote:
 Originally Posted by CRGreathouse I don't read German...
Read this then : http://math.eretrandre.org/tetrationfor ... hp?tid=771

Trappman explains Kneser's method excellently and even for a non-professional like me can understand it completely.

April 29th, 2013, 07:03 AM   #6
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Re: Limit involving 1/2-height superexponential

Quote:
 Originally Posted by mathbalarka $\frac{1}{n} \sum_{j=1}^{n} \sum_{i=1}^{j} \frac{1}{{}^{1/2} i}$
Using the fact that $\mathrm{sexp}(i, 1/2) \leq i^{1/2}$, we can bound up the sum above by $\frac{1}{n} \sum_{j= 1}^{n} \sum_{i = 1}^{j} i^{-1/2}$ which grows like o(n) forcing to conclude that the original sum grows as o(n) too.

EDIT : The calculations above have errors since the sum $\frac{1}{n} \sum_{j= 1}^{n} \sum_{i = 1}^{j} i^{-1/2}$ isn't really an upper bound for the one of interest, instead, it's a lower bound. Which means, proving that $\lim_{n \to \infty} \frac{1}{n} \sum_{j= 1}^{n} \sum_{i = 1}^{j} i^{-1/2}$ diverges would prove that $\lim_{n \to \infty} \frac{1}{n} \sum_{j=1}^{n} \sum_{i=1}^{j} \frac{1}{{}^{1/2} i}$ diverges. So, the problem is being reduced to a fairly easy Real analysis question, does $\lim_{n \to \infty} \frac{1}{n} \sum_{j= 1}^{n} \sum_{i = 1}^{j} i^{-1/2}$ converges or diverges? I know it's likely that this will diverge since $\zeta$$\frac{1}{2}$$$ is not Cesaro summable, but a rigorous proof would be nice to see.

Balarka
.

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