If-a-lt-1-and-b-lt-1-then-the-sum-of-the-series-1-1-a-b-1-a-a-2-b-2-1-a-a-2-a-3-b-3-is-

Question Number 68151 by mhmd last updated on 06/Sep/19

If ∣ a ∣< 1 and ∣ b ∣< 1, then the sum of the series

1 + (1 + a) b + (1 + a + a^{2}) b^{2} + (1 + a + a^{2} + a^{3}) b^{3} + \dots

is

Commented by ~ À ® @ 237 ~ last updated on 06/Sep/19

L e t i t b e S . W e c a n a s c e r t a i n t h a t

S = \underset{n = 0}{\sum^{\infty}} (1 + a + a^{2} + \dots . . + a^{n}) b^{n}

k n o w i n g t h a t 1 + a + \dots + a^{n} = \frac{1 - a^{n + 1}}{1 - a}

S = \underset{n = 0}{\sum^{\infty}} (\frac{1 - a^{n + 1}}{1 - a}) b^{n} = \frac{1}{1 - a} (\underset{n = 0}{\sum^{\infty}} b^{n} - a \underset{n = 0}{\sum^{\infty}} {(a b)}^{n})

U s i n g \frac{1}{1 - t} = \underset{n = 0}{\sum^{\infty}} t^{n}

S = \frac{1}{1 - a} (\frac{1}{1 - b} - \frac{a}{1 - a b}) = \frac{1}{(1 - b) (1 - a b)}