Power series and power expansion with absolute convergence.

Question

Suppose $|a|+|b|<1$ and absolutely convergent. How can I show $$ \frac{1}{1-a-b} = \sum_{n=0}^\infty \frac{a^n}{(1-b)^{n+1}}? $$

I tried to start from here: $$ \frac{1}{1-a-b}=\frac{1}{1-(a+b)}=\sum_{m,n}{m+n\choose n}a^nb^m $$ then I tried to use the fact that we have absolute convergence but haven't gone nowhere.

Answer 1

HINT

One possible direction to look at is $$ \sum_{n=0}^\infty \frac{a^n}{(1-b)^{n+1}} = \frac{1}{1-b} \sum_{n=0}^\infty \frac{a^n}{(1-b)^n} = \frac{1}{1-b} \sum_{n=0}^\infty \left(\frac{a}{1-b}\right)^n $$ and you can use the geometric series on the sum for $|a| < |1-b|$.

Answer 2

Another hint: using $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$ for $|x|<1$ solves the problem.