MHB Limit of $(x_{n})_{n\geq 1} with Given Conditions

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The sequence defined as \( x_n = ac + (a + ab)c^2 + \ldots + (a + ab + \ldots + ab^n)c^{n+1} \) converges to the limit \( \frac{ac}{(1-bc)(1-c)} \) under the conditions \( |c| < 1 \), \( b \neq 1 \), and \( |bc| < 1 \). The discussion highlights the importance of recognizing the sums as geometric progressions, which start from 1. The correct formulation involves adjusting the index of summation to accurately capture the series. The participant initially doubted their calculations but ultimately confirmed their solution was correct after reviewing the steps. Clarification on the setup of the sums was key to reaching the correct limit.
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Hi!

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 $x_{n}$.

My attempt is in the picture.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?
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I believe the correct set up is $\displaystyle x_n = \sum_{j=1}^{n+1}\bigg(\sum_{k=0}^{j-1}ab^k\bigg)c^j = \sum_{j=1}^{n+1}\frac{a(b^j-1)}{b-1}c^j$ so that $\displaystyle \lim_{ n \to \infty} x_n = \sum_{j=1}^{\infty}\frac{a(b^j-1)}{b-1}c^j = \frac{ac}{(c-1)(bc-1)}.$
 
I'm stupid, I got the correct answer.I just needed to solve some little calculations.I don't know I thought I'm wrong..
Thanks!
 
Vali said:
I'm stupid, I got the correct answer.I just needed to solve some little calculations.I don't know I thought I'm wrong..
Thanks!
I see. I didn't go through your calculations because I couldn't zoom in tbh.
 
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