From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Re : Cauchy and d'Alembert sequences !!!
Date: 14 Nov 1995 18:40:01 GMT

In article <ejouvent-1311950955560001@ts4-p18.dialup.iway.fr>,
Eric Jouvent <ejouvent@pratique.fr> wrote:
>I'm a second year student in physics in Paris and I have some difficulties
>to solve this problem :
>
>U(n+1)/U(n) = 1 - a/n + V(n)
>
>with a>0
>and … [abs(V(n))] is convergent

(There is a non-ASCII character (\205) here -- is it a capital sigma? I assume
the poster means to say  Sum | V(n) |  converges.)

>I have to proove that it exists a positive real number K so that :
>U(n) is equivalent to K/(npower(a)).

This is pretty standard. As a physics student you would, I imagine, have to
do this kind of thing frequently.

Let  u(n)=log U(n). Then you have
	u(n+1)-u(n) = log(1 - a/n + V(n)) = -a/n + V(n) + H(n)
where  H(n) includes all the terms in  log(1+x)-x = -x^2/2+x^3/3+... (The
statement that the  Sum |V(n)| converges implies most  -a/n+V(n) are less
than  1  in magnitude, making this Taylor series converge.) Summing, we
have  u(N)=u(1)- a Sum(1/n, n<N) + Sum( V(n), n<N ) + Sum( H(n), n<N ),
and thus  u(N) = -a log(N) + Sum( H(n), n<N ) + O(1). All you need is some
good estimates on the  H(n)  to keep this last sum also finite.

Here's an approach that will work. For small  x  we can estimate
-C1 x^2 < log(1+x)-x < -C2 x^2  (for example, if  |x| < 0.5, we can use
C1=0.78, C2=0.37). In particular, for almost all  n  we have  H(n) negative
and of magnitude  |H(n)| < C1 (a/n-V(n))^2. Thus the sum of the  H(n)
consists of three parts:
	Sum( 1/n^2, n<N )  is known to converge  (to  pi^2/6).
	Sum( V(n)/n, n<N ) is in value less than  Sum( |V(n)|, n<N), and
so converges.
	Sum( V(n)^2, n<N ) is likewise less than  Sum( |V(n)|, n<N)  except
for at most a finite number of summands in which  |V(n)| > 1.

We conclude that the sums  Sum( H(n), n< N )  also stay bounded as  N  
increases, so that u(N) = -a log(N) + O(1), and thus there are constants
with    K1 N^a < U(N) < K2 N^a.

A slightly more careful analysis gives the better result  
	u(N) = -a log(N) + C + o(1)
for some constant  C, so that there is a constant  K  such that the ratios
	U(N) / (K/N^a)
approach  1  as  N --> oo. 

dave