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Thread: How can you prove this?

  1. #1
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    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;

    $\displaystyle \sum_{n=1}^{{\color{red}\infty}}r^n = \frac{r}{1-r}$
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  2. #2
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    Hint: $\displaystyle r^{n}=\frac{r^{n}-r^{n+1}}{1-r},$ and use telescoping series.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    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;

    $\displaystyle r^{n}=\frac{r^{n}-r^{n+1}}{1-r}$
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  4. #4
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    $\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} $
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  5. #5
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    Quote Originally Posted by Plato View Post
    $\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.
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  6. #6
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    Quote Originally Posted by Plato View Post
    $\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?
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  7. #7
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    Quote Originally Posted by RhysGM View Post
    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=
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  8. #8
    MHF Contributor Drexel28's Avatar
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    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.
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    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.
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