From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: test for rational sereies convergence??
Date: 9 Sep 1995 04:16:37 GMT
In article ,
keller@techunix.technion.ac.il (Keller Yuli) wrote:
>
> How can i tell whether a rational number series converges to a
> rational/irrational number??
to which Gerry Myerson correctly responded
>If you can figure this one out, you will be a very famous person.
(and indeed many numbers of unknown rational status can be represented as
a rational series).
That said, you can sometimes show a series converges to an irrational
number by noting that it converges too quickly. For example, let
x = Sum ( 1/n! ) = 2.71828... The n-th partial sum is a_n/n!, say,
but differs from x itself by, oh, about 1/(n * n!) (actually the
difference is less than this). The key fact here is that the errors
go to zero distinctly faster than than the denominators are growing.
So _if_ x were rational, say x = p/q, then
| x - (a_n/n!) | = | p * n! - a_n * q | / (q * n!)
would have to be at least 1/( q * n! ) (the difference can't be
zero since every partial sum is strictly less than x). So now you're
stuck with the inequalities
1/(q * n!) <= error < 1/(n * n!),
which can't be true for all n (it's not true for n > q). This gives
a contradiction, showing that x is irrational.
This is really lesson 1 in transcendental number theory. There's a
lot you can say by looking at the speed of convergence of rational
approximations, e.g. resulting from continued fractions, applied to
algebraic numbers, and so on. I suppose Baker's book is the usual reference.
dave