limit with sum and geometric progression I have the following sequence $$(x_{n})_{n\geq 1}, \ x_{n}=ac+(a+ab)c^{2}+...+(a+ab+...+ab^{n})c^{n+1}$$
Also I know that $a,b,c\in \mathbb{R}$ and $|c|<1,\ b\neq 1, \ |bc|<1$
I need to find the limit of $x_{n}$.
The result should be $\frac{ac}{(1-bc)(1-c)}$
I miss something at these two sums which are geometric progressions.Each sum should start with $1$ but why ? If k starts from 0 results the first terms are $bc$ and $c$ right?
My attempt:
$x_{n}=a(c+c^{2}(1+b)+...+c^{n+1}(1+b+...+b^{n}))$
$1+b+...+b^{n}=\frac{b^{n+1}-1}{b-1}$ so $$x_{n}=a\sum_{k=0}^{n}c^{k+1}\cdot \frac{b^{k+1}-1}{b-1}\Rightarrow x_{n}=\frac{a}{b-1}\sum_{k=0}^{n}c^{k+1}\cdot (b^{k+1}-1)=\frac{a}{b-1}(\sum_{k=0}^{n}c^{k+1}\cdot b^{k+1}-\sum_{k=0}^{n}c^{k+1})$$
Now I take separately each sum to calculate.
$\sum_{k=0}^{n}(bc)^{k+1}=bc+b^2c^2+...+b^{n+1}c^{n+1}$
It's a geometric progression with $r=bc$, right ?But if a calculate the sum, in the end I don't get the right answer.I get the right answer if this progression starts with $1$ as first term.Why?
The same thing with the second sum.If the first term is $1$ I'll get the right answer.
Why I need to add/subtract a $1$ to get the answer?Why I don't get the correct answer just by solving the progressions with the first term $bc$ and $c$?
 A: You seem to be doing everything correctly. Using your final value for $x_n$, and taking the limit as $n \to \infty$, I get, using the sum of an infinite geometric series being $\frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio where $\left|r\right| \lt 1$, of
$$\cfrac{a}{b-1}\left(\cfrac{bc}{1-bc} - \cfrac{c}{1-c}\right) \tag{1}\label{eq1}$$
For the part inside the brackets, multiply the first term's numerator & denominator by $1-c$ and the first term's numerator & denominator by $1-bc$, to get a common denominator, with this then becoming
$$\cfrac{bc - bc^2 - c + bc^2}{\left(1-bc\right)\left(1-c\right)}$$
$$\cfrac{c\left(b-1\right)}{\left(1-bc\right)\left(1-c\right)} \tag{2}\label{eq2}$$
Substituting this into \eqref{eq1}, then removing the common factor of $b - 1$ (as $b \neq 1$) gives your expected result of
$$\cfrac{ac}{\left(1-bc\right)\left(1-c\right)} \tag{3}\label{eq3}$$
A: Hint: Take $ac$ common. For finding the limit of the sequence, consider $n \rightarrow \infty$ and apply the formula for summation of geometric series with infinite terms. Since your $r<1$, $S = \frac{a}{1-r}$ holds.
