Can someone explain why this is a proof?

  • Thread starter flyingpig
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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.
 
  • #3
34,017
5,672
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]
 
  • #4
2,571
1
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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