On Feb 24, 4:50�pm, Steven D'Aprano <[EMAIL PROTECTED] cybersource.com.au> wrote: > On Sun, 24 Feb 2008 11:09:32 -0800, Lie wrote: > > I decided to keep the num/den limit low (10) because higher values might > > obscure the fact that it do have limits. > > You do realise that by putting limits on the denominator, you guarantee > that the sum of the fractions also has a limit on the denominator? In > other words, your "test" is useless. > > With denominators limited to 1 through 9 inclusive, the sum will have a > denominator of 2*3*5*7 = 210.
Th limit will be 2*2*2*3*3*5*7. As MD said, "equivalently the product over all primes p <= n of the highest power of p not exceeding n". > But that limit is a product (literally and > figuratively) of your artificial limit on the denominator. Add a fraction > with denominator 11, and the sum now has a denominator of 2310; add > another fraction n/13 and the sum goes to m/30030; and so on. > > -- > Steven -- http://mail.python.org/mailman/listinfo/python-list
