Actually, every rational number has a recurring cycle in a positional representation ... it's just that for some numbers it is a trivial "0" cycle. But that's just a quibble.

Assuming that the provided string is long enough for two cycles to appear gets boring pretty quickly ... on my system that assumption fails already for 1/17, unless I use bignum. So I'll do that. And, true to habit, I'll use a regex to solve the problem. :)

(Pass an argument to the program to get the cycle for the inverse of more (or less) than the default 99 first natural numbers:)

#!/usr/bin/perl

use strict;
use warnings;

use constant COUNT  => $ARGV[0]||99;
use constant LENGTH => length COUNT;

use bignum a => 3*COUNT; # accuracy ...

for my $n (1..COUNT) {
  my $f = substr(1/$n, 0, 3*COUNT); # precision too ...
  if ( $f =~ /(\d+?)\1+(?!\d{${\COUNT}})/ ) { # a little leeway ...
    printf " 1/%-*d = %-30.30s -> %s\n", LENGTH, $n, $f, $1;
  }
  else {
    # should never happen ...
    die "Insufficient precision/accuracy?  Cannot handle the following
+:\n$f\n";
  }
}
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Thank you — that was an interesting distraction. :)

Update: Made the leeway big enough that we get the cycle sequence as it first appears, and not some arbitrary cycling of it, no matter how long the sequence.