a-gt-0-and-b-gt-0-if-1-1-ax-1-bx-n-a-n-x-n-find-the-sequence-a-n-

Question Number 29970 by abdo imad last updated on 14/Feb/18

a > 0 a n d b > 0 i f \frac{1}{(1 - a x) (1 - b x)} = \sum_{n} a_{n} x^{n}

f i n d t h e s e q u e n c e a_{n} .

Commented by abdo imad last updated on 16/Feb/18

f o r ∣ x ∣< i n f (\frac{1}{∣ a ∣}, \frac{1}{∣ b ∣}) w e h a v e

\frac{1}{(1 - a x) (1 - b x)} = (\sum_{n = 0}^{\infty} a^{n} x^{n}) (\sum_{m = 0}^{\infty} b^{m} x^{m})

= \sum_{k = 0}^{\infty} c_{k} x^{k} w i t h c_{k} = \sum_{i + j = k} u_{i} v_{j} = \sum_{i + j = k} a^{i} b^{j}

= \sum_{i = 0}^{k} a^{i} b^{k - i} = \frac{a^{k + 1} - b^{k + 1}}{a - b} i f a \neq b a n d i f a = b

c_{k} = \sum_{i = 0}^{k} a^{i} = \frac{1 - a^{k + 1}}{1 - a} i f a \neq 1 a n d c_{k} = (k + 1) i f a = 1 f i n a l l y

i f a \neq b a_{n} = \frac{a^{n + 1} - b^{n + 1}}{a - b} a n f i f a = b \neq 1 a_{n} = \frac{1 - a^{n + 1}}{1 - a}

i f a = b = 1 a_{n} = n + 1 a n d w e g e t t h e f o r m u l a

\frac{1}{{(1 - x)}^{2}} = \sum_{n = 0}^{\infty} (n + 1) x^{n} .