[jason_t]

The sum of any series of the form sum from n = 0 to infinity of x^n is 1/(1-x) as long as |x| < 1.

So the sum from n = 1 to infinity of numerator * 7 * (2/100)^n is (numerator) * 7 * (1/(1-2/100) - 1) = numerator * 7 / 49 = numerator / 7.

In general, numerator / d = sum from n = 1 to infinity of numerator * d * [1/(d^2+1)]^k, as you can work out using the above formula.

as for the formula, here is the reason this is true

let s = 1 + x + x^2 + ... + x^n
then x * s = x + x^2 + ... + x^(n+1)

so x * s - s = x^(n+1) - 1

and therefore

(x-1) * s = x^(n+1) - 1

so

s = [ (x^(n+1) - 1) / (x - 1) ]

and if |x| < 1 then letting n->infinty we have x^(n+1)->0 so s->-1/(x-1) = 1/(1-x).