Can someone explain why this is a proof?

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Homework Help Overview

The discussion revolves around proving the convergence of the geometric series \(\sum_{n=1}^{\infty} r^n\) for values of \(r\) in the range \(-1 < r < 1\). Participants are examining the implications of a limit derived from the partial sums of the series.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how the limit of the partial sums, \(\lim_{n\to\infty}s_n=\frac{r}{1-r}\), serves as a proof of convergence for the series. Some participants provide definitions related to convergence and question the accuracy of the original poster's statements regarding limits.

Discussion Status

The discussion is active, with participants providing clarifications on the definition of convergence and correcting a misunderstanding regarding the behavior of \(r^{n+1}\) as \(n\) approaches infinity. There is no explicit consensus, but guidance has been offered regarding the correct interpretation of limits in the context of series convergence.

Contextual Notes

Participants are addressing a common misunderstanding related to the behavior of terms in the series as \(n\) increases, specifically focusing on the limit of \(r^{n+1}\) and its implications for convergence.

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Homework Statement



Prove that the geometric series [tex]\sum_{n=1}^{\infty} r^n[/tex] if -1 < r < 1


2. The Solution

[tex]s_n = r + r^2 + ... + r^n[/tex]

[tex]rs_n =r^2 + r^3 ... + r^{n+1}[/tex]

[tex]s_n - rs_n = r - r^{n+1}[/tex]

[tex]s_n = \frac{r - r^{n+1}}{1 -r}[/tex]

For |r|<1

[tex]As\;n\to\infty\;,r^{n+1}\to \infty[/tex]

Therefore

[tex]\lim_{n\to\infty}s_n=\frac{r}{1-r}[/tex]

Q.E.D

Question

The solution is what we took in notes during lecture.

Now here is my question why does [tex]\lim_{n\to\infty}s_n=\frac{r}{1-r}[/tex] answer the proof? How does that prove the geometric series [tex]\sum_{n=1}^{\infty} r^n[/tex] converge?
 
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Because s_n is a partial sum of the series. And the definition of convergence of a series is that the limit of the partial sums converges. Look up the definition of convergence.
 
There is an error in what you wrote.
|r| < 1
[tex]As\;n\to\infty\;,r^{n+1}\to \infty[/tex]
You should have
[tex]As\;n\to\infty\;,r^{n+1}\to 0[/tex]
 
Mark44 said:
There is an error in what you wrote.

You should have
[tex]As\;n\to\infty\;,r^{n+1}\to 0[/tex]

Yeah lol, it seems I make one of these TEX mistakes every time lol

thanks for catching that
 

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