(sum_{k=0 to n} r^k)(r-1) = sum_{k=0 to n} [r^{k+1} - r^k]

.....................................= (r-1) + (r^2-r) + ... + (r^n-r^{n-1)) + (r^{n+1} - r^n)

Now all but the -1 in the first term and the r^{n+1} in the last term cancel

so:

(sum_{k=0 to n} r^k)(r-1) = [r^{n+1}-1]

or:

sum_{k=0 to n} r^k = [r^{n+1}-1]/(r-1).

From the first part we have:then I'm supposed to use part to show why

the sum K=0 to infinity = (1)/(1-r)

(sum_{k=0 to n} r^k) = [r^{n+1}-1]/(r-1)

Now if |r|<1

lim_{n->infty} (sum_{k=0 to n} r^k) = (sum_{k=0 to infty} r^k) = -1/(r-1) = 1/(1-r)

as lim_{n->infty} r^{n+1} = 0.

RonL