- #1
flyingpig
- 2,579
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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?