How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?    
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance! 

Let $$S = 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$  
Also we know, $$(1-b)(1+b+b^2+b^3 \cdots +b^n-1) = 1-b^n$$ 
We now start feeling that having $(1-b)$ as a factor in each term in the $RHS$ will create a "nice" series.
So multiplying the $LHS$ & $RHS$ by $(1-b)$, we get,        $$S(1-b) = (1-b)+a(1-b^2)+a^2(1-b^3)+a^3(1-b^4)+\cdots $$
Now, upon expanding the $RHS$, we get 2 geometric progressions,
     $$S(1-b) = (1+a+a^2+a^3+\cdots)-(b+ab^2+a^2b^3+a^3b^4+\cdots)$$ 
The first series is an infinite  GP ( here $|a|<1$ is a constraint) with the first term equal to $1$ and common ratio $a$. So its sum, $S_1$ (say) is given by, 
      $$S_1 = \frac{1}{1-a}$$ 
The second series is also an infinite GP ( here too $|ab|<1$ is a constraint). It's first term is $b$ and the common ratio is $ab$. It's sum, $S_2$ (say), is given by,
      $$S_2=\frac{b}{1-ab}$$
Combining $S_1$ and $S_2$ we finally get,
      $$S(1-b) = \frac{1}{1-a} - \frac{b}{1-ab}$$
Simplifying the $RHS$ and dividing both sides of the equation by $(1-b)$ we get,
                              $$S = \frac{1}{(1-a)(1-ab)}$$
