# How can you prove this?

• Jan 6th 2010, 08:19 AM
RhysGM
How can you prove this?
$\displaystyle \sum_{n=1}^{{\color{red}\infty}}r^n$

I'm assuming this is the correct way to write;

$\displaystyle r+r^2+r^3+r^4 . . . r^n$ where the n goes on for infinity.

So how can you prove;

$\displaystyle \sum_{n=1}^{{\color{red}\infty}}r^n = \frac{r}{1-r}$
• Jan 6th 2010, 08:30 AM
Krizalid
Hint: $\displaystyle r^{n}=\frac{r^{n}-r^{n+1}}{1-r},$ and use telescoping series.
• Jan 6th 2010, 10:18 AM
RhysGM
Quote:

Originally Posted by Krizalid
Hint: $\displaystyle r^{n}=\frac{r^{n}-r^{n+1}}{1-r},$ and use telescoping series.

Have I understood this correctly;

$\displaystyle \frac{r^{n}-r^{n+1}}{1-r} = \frac{r^{n}}{1-r} + \frac{-r^{n+1}}{1-r}$

$\displaystyle = \left(\frac{r}{1-r} + \frac{-r^{2}}{1-r}\right) + \left(\frac{r^{2}}{1-r} + \frac{-r^{3}}{1-r}\right) + \left(\frac{r^{3}}{1-r} + \frac{-r^{4}}{1-r}\right) + . . .$

$\displaystyle = \frac{r}{1-r} + \left(\frac{-r^{2}}{1-r}+\frac{r^{2}}{1-r}\right) + \left(\frac{-r^{3}}{1-r} + \frac{r^{3}}{1-r}\right) + . . .$

$\displaystyle = \frac{r}{1-r}$

If that is correct how did you get;

$\displaystyle r^{n}=\frac{r^{n}-r^{n+1}}{1-r}$
• Jan 6th 2010, 10:49 AM
Plato
$\displaystyle \begin{gathered} S = r + r^2 + \cdots + r^n \hfill \\ rS = r^2 + r^3 + \cdots + r^{n + 1} \hfill \\ S - rS = r - r^{n + 1} \hfill \\ S = \frac{{r - r^{n + 1} }} {{1 - r}} \hfill \\ \end{gathered}$

Now $\displaystyle \begin{gathered} \left| r \right| < 1\quad \Rightarrow \quad \lim _{n \to \infty } r^{n + 1} = 0 \hfill \\ S \to \frac{r} {{1 - r}} \hfill \\ \end{gathered}$
• Jan 6th 2010, 10:57 AM
RhysGM
Quote:

Originally Posted by Plato
$\displaystyle \begin{gathered} S = r + r^2 + \cdots + r^n \hfill \\ rS = r^2 + r^3 + \cdots + r^{n + 1} \hfill \\ S - rS = r - r^{n + 1} \hfill \\ S = \frac{{r - r^{n + 1} }} {{1 - r}} \hfill \\ \end{gathered}$

Now $\displaystyle \begin{gathered} \left| r \right| < 1\quad \Rightarrow \quad \lim _{n \to \infty } r^{n + 1} = 0 \hfill \\ S \to \frac{r} {{1 - r}} \hfill \\ \end{gathered}$

awesome, cheers.
• Jan 7th 2010, 01:05 AM
RhysGM
Quote:

Originally Posted by Plato
$\displaystyle \begin{gathered} S = r + r^2 + \cdots + r^n \hfill \\ rS = r^2 + r^3 + \cdots + r^{n + 1} \hfill \\ S - rS = r - r^{n + 1} \hfill \\ S = \frac{{r - r^{n + 1} }} {{1 - r}} \hfill \\ \end{gathered}$

Now $\displaystyle \begin{gathered} \left| r \right| < 1\quad \Rightarrow \quad \lim _{n \to \infty } r^{n + 1} = 0 \hfill \\ S \to \frac{r} {{1 - r}} \hfill \\ \end{gathered}$

What do I google to get more explanation on this? What is this method called?
• Jan 7th 2010, 02:18 AM
mr fantastic
Quote:

Originally Posted by RhysGM
What do I google to get more explanation on this? What is this method called?

sum infinite geometric series: sum infinite geometric series - Google Search=
• Jan 7th 2010, 01:39 PM
Drexel28
I gave you this explanation in the original thread.

As for Krizalid's answer merely note that $\displaystyle r^n=r^n\cdot\frac{r-1}{r-1}=\frac{r^{n+1}-r^n}{1-r}$. So, for a finite sum $\displaystyle \sum_{k=1}^{n}r^k=\sum_{k=1}^n\frac{r^{k+1}-r^k}{r-1}=\frac{1}{r-1}\sum_{k=1}^{n}\left\{r^{k+1}-r^k\right\}$ and we notice that the sum is the same as $\displaystyle \left(r^2-r\right)+\left(r^3-r^2\right)+\cdots\left(r^{n+1}-r^{n}\right)$ which simplifies to $\displaystyle r^{n+1}-r$. Taking the limit $\displaystyle n\to\infty$ finishes the argument.
• Jan 13th 2010, 04:00 AM
RhysGM
Quote:

Originally Posted by Drexel28
I gave you this explanation in the original thread.

I know, I didn't understand it then, I needed to do more research before I could understand what you were telling me.