a(1) = s(1) = 1 + (r-1)a(1),a(1) = 1/(2-r).
s(n) = 1 + (r-1)a(n),
r = 1時,s(n) = 1,a(1) = 1,a(n) = 1.
lim_{n->無窮｝s(n) = 1,滿足題意.
r不為1時,
s(n+1) = 1 + (r-1)a(n+1),
a(n+1) = s(n+1) - s(n) = (r-1)a(n+1) - (r-1)a(n),
a(n+1) = [(1-r)/(2-r)]a(n),
{a(n)}是首項為a(1)=1/(2-r),公比為(1-r)/(2-r)的等比數(shù)列.
a(n) = [1/(2-r)][(1-r)/(2-r)]^(n-1),
s(n) = 1 + (1-r)a(n) = 1 + [(1-r)/(2-r)]^n,
0 = lim_{n->無窮｝[(1-r)/(2-r)]^n,
|(1-r)/(2-r)| < 1,
|1-r| < |2-r|,
1 - 2r + r^2 < 4 - 4r + r^2,
2r < 3,
r < 3/2.
綜合,有,
r的取值范圍為,r< 3/2