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! 
 A: HINT:
$$\sum_{n=0}^\infty a^n\sum_{m=0}^n b^m=\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}$$
Can you finish?
A: Since the OP admitted that they can't continue from Dr.MV's hint, I'll do it myself:
$$\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}=\frac{1}{1-b} \left(\sum_{n=0}^\infty a^n-b \sum_{n=0}^\infty (ab)^n \right)=$$
Here we use geometric series, so we need to have $|a|<1$ and $|ab|<1$: *
$$=\frac{1}{1-b} \left(\frac{1}{1-a}-\frac{b}{1-ab} \right)=\frac{1}{(1-a)(1-ab)}$$
* Correction due to Batominovski's comment
A: 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,
                              $$S = \frac{1}{(1-a)(1-ab)}$$
A: A slight variant that requires no explicit cancelling of $1-b$ (so you needn't handle the $b=1$ case with a continuity argument) rearranges the double summation, viz. $$\sum_{n=0}^\infty\left(a^n\sum_{k=0}^n b^k\right)=\sum_{k=0}^\infty b^k\sum_{n=k}^\infty a^n=\sum_{k=0}^\infty b^ka^k\frac{1}{1-a}=\frac{1}{\left( 1-a\right)\left( 1-ab\right)}$$ provided $\left| a\right| < 1$ and $\left| ab\right| <1$.
A: You have $$\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)\text{.}$$ Before determining what this converges to, it is worth establishing for which values of $(a,b)$ it converges at all. To that end, consider the Ratio Test:
$$\left\lvert\frac{a^{n+1}\sum_{m=0}^{n+1}b^m}{a^n\sum_{m=0}^nb^m}\right\rvert=\left\lvert a\left(1+\frac{b^{n+1}}{\sum_{m=0}^nb^m}\right)\right\rvert\to\begin{cases}\lvert ab\rvert&\lvert b\rvert>1\\\lvert a\rvert&\lvert b\rvert=1\\\lvert a\rvert&\lvert b\rvert<1\\\end{cases}$$
So it converges when:


*

*$\lvert ab\rvert<1$ and $\lvert b\rvert>1$

*$\lvert a\rvert<1$ and $\lvert b\rvert\leq1$


And diverges when:


*

*$\lvert ab\rvert>1$ and $\lvert b\rvert>1$

*$\lvert a\rvert>1$ and $\lvert b\rvert\leq1$


This test has been inconclusive when:


*

*$\lvert ab\rvert=1$ and $\lvert b\rvert>1$

*$\lvert a\rvert=1$ and $\lvert b\rvert\leq1$


Let's look at the last case first. If $\lvert a\rvert=1$, then you have $\sum_{n=0}^\infty {\pm_n\left(\sum_{m=0}^nb^m\right)}\text{.}$ The terms of such a series, $\pm_n\left(\sum_{m=0}^nb^m\right)$,  do not approach $0$ no matter what $b$ is. So your series will diverge. 
What if $\lvert b\rvert>1$ and $\lvert a\rvert=\frac{1}{\lvert b\rvert}$? Since $\lvert b\rvert>1$ we may write $\sum_{m=0}^nb^m=\frac{b^{n+1}-1}{b-1}$, and your series is $\sum_{n=0}^\infty {\pm_n\left(\frac{b^{n+1}-1}{b^n(b-1)}\right)}\text{.}$ Again using $\lvert b\rvert>1$, the terms of this series do not converge to $0$, so this series would be divergent. 
Now we know the only situations where your series converges are:


*

*$\lvert ab\rvert<1$ and $\lvert b\rvert>1$ (which implies $\lvert a\rvert<1$)

*$\lvert a\rvert<1$ and $\lvert b\rvert\leq1$ (which implies $\lvert ab\rvert<1$)


When $b\neq1$, both situations give:
$$\begin{align}
\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)
&=\sum_{n=0}^\infty a^n\frac{1-b^{n+1}}{1-b}\\
&=\frac{1}{1-b}\left(\sum_{n=0}^\infty a^n-b\sum_{n=0}^\infty (ab)^n\right)\\
&=\frac{1}{1-b}\left(\frac{1}{1-a}-\frac{b}{1-ab}\right)\\
&=\frac{1}{(1-a)(1-ab)}\\
\end{align}$$
And when $b=1$ with $\lvert a\rvert<1$,
$$\begin{align}
\sum_{n=0}^\infty \left(a^n\sum_{m=0}^nb^m\right)
&=\sum_{n=0}^\infty (n+1)a^n\\
&=\left.\sum_{n=0}^\infty \frac{d}{dx}x^{n+1}\right|_{x=a}\\
&=\left.\frac{d}{dx}\sum_{n=0}^\infty x^{n+1}\right|_{x=a}\\
&=\left.\frac{d}{dx}\left(\frac{1}{1-x}-1\right)\right|_{x=a}\\
&=\left.\frac{1}{(1-x)^2}\right|_{x=a}\\
&=\frac{1}{(1-a)^2}\\
\end{align}$$
Which conveniently agrees with the formula $\frac{1}{(1-a)(1-ab)}$.
