NetNews Usenet Archive 1992 #16/NN_1992_16.iso/spool/sci/math/9608

Text File | 1992-07-28 | 2.4 KB | 63 lines

Newsgroups: sci.math
Path: sparky!uunet!usc!cs.utexas.edu!sun-barr!ames!pasteur!bellini.berkeley.edu!luzeaux
From: [e-mail] (Dominique Luzeaux)
Subject: Re: Integer Sequence Problem
Message-ID: <[e-mail]>
Sender: [e-mail] (NNTP Poster)
Nntp-Posting-Host: bellini.berkeley.edu
Reply-To: [e-mail] (Dominique Luzeaux)
Organization: University of California, Berkeley
References:  <[e-mail]>
Date: Tue, 28 Jul 1992 22:56:05 GMT
Lines: 49

In article <[e-mail]>,
[e-mail] (Hannes Hassler) writes:
|> For S an infinite, increasing sequence of positive integers (1=a_1 < a_2 <
|> a_3 < ... < a_i < a_{i+1} < ...), we define the VALUE val(S) of S
|>
|>         val(S)  :=    sup_{n>1}  \sum_{i=1}^n a_i   / a_{n-1}
|>
|> (in words: For every n, sum up the first n terms of S and divide this sum
|> by the (n-1)st term.  Then val(S) is the supremum of all these values).
|>
|> PROBLEM: Find the minimum (infimum) of val(S) over all possible sequences S.

Let us call : s_n=\frac{\sum_{i=1}^n}a_i}{a_{n-1}}
and : q_n=\frac{a_n}{a_{n-1}}

Then : s_n - s_{n-1} = q_n - (1-1/q_{n-1}) s_{n-1}
Let us suppose val(S) is finite for some sequence S; in other words val(S)
is an adherence value of (s_n); we can then consider that s_n converges to
val(S) (it is always possible to find a subsequence converging to val(S) and
we consider then this particular subsequence).

If s_n converges to some limit L, this difference converges to 0; this
implies that:
        q_n
     -------------  --> L
     1 - 1/q_{n-1}
which can be rewritten:  q_n = (L + e_n)(1 - 1/q_{n-1})  where e_n is a small
perturbation.
Draw the function x \mapsto (L+e)(1-1/q_{n-1}) where e is small; after some
iterations, q_n will be negative which is impossible by definition of S; this
holds even if e is allowed to change for each n, but remains in a bounded small
interval. The only way to avoid that situation is to have q_n = q_{n-1}
starting
from some n; let q be this constant value. As q has to be integer and
greater or
equal than 2, let us try q=2. This yields L=4.

In fact, the sequence (1,2,...,2^n,...) satisfies all the conditions and yields
val(S)=4

Furthermore, we can say that val(S) takes as only finite values q^2/(q-1)
where q is an integer >=2, hence 4 is the minimum of val(S) over all
sequences S.


    Dominique Luzeaux