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