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

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Discussion Overview

The discussion revolves around the limit of a sequence defined by a specific formula involving parameters \(a\), \(b\), and \(c\) under certain conditions. Participants explore the setup of the sequence and the calculations needed to derive its limit, focusing on the nature of geometric series involved.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant presents the sequence \(x_n\) and expresses uncertainty about the correct setup of the sums involved, particularly regarding the starting index of the geometric series.
  • Another participant proposes a different formulation for \(x_n\) using a double summation and provides a limit expression, suggesting that the limit can be expressed as \(\frac{ac}{(c-1)(bc-1)}\).
  • A later reply indicates that the initial participant resolves their confusion and arrives at the correct answer after further calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial formulation of the sequence, but there is acknowledgment of a correct answer by one participant after further reflection. The discussion includes both uncertainty and resolution of confusion.

Contextual Notes

There are unresolved aspects regarding the setup of the sums and the conditions under which the limits are derived, particularly the implications of the parameters \(a\), \(b\), and \(c\) and their relationships.

Vali
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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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