# Math Help - How can you prove this?

1. ## How can you prove this?

$\sum_{n=1}^{{\color{red}\infty}}r^n$

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

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

So how can you prove;

$\sum_{n=1}^{{\color{red}\infty}}r^n = \frac{r}{1-r}$

2. Hint: $r^{n}=\frac{r^{n}-r^{n+1}}{1-r},$ and use telescoping series.

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

Have I understood this correctly;

$

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

$
= \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) + . . .
$

$
= \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) + . . .
$

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

If that is correct how did you get;

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

4. $\begin{gathered}
S = r + r^2 + \cdots + r^n \hfill \\
rS = r^2 + r^3 + \cdots + r^{n + 1} \hfill \\
S - rS = r - r^{n + 1} \hfill \\
S = \frac{{r - r^{n + 1} }}
{{1 - r}} \hfill \\
\end{gathered}$

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

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

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

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

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

What do I google to get more explanation on this? What is this method called?

7. 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=

8. I gave you this explanation in the original thread.

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

9. 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.