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Thread: Series summation doubt

  1. #1
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    Series summation doubt

    Hi there. I have some doubts with this exercise. I have to make the summation for:

    $\displaystyle \displaystyle\sum_{i=-3}^\infty{(-\displaystyle\frac{1}{3})^i}$

    I've solved this way:

    $\displaystyle \displaystyle\sum_{i=-3}^{\infty}{(-\displaystyle\frac{1}{3})^i}=\displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}+\displaystyle\sum_{i= 1}^{\infty}{(-\displaystyle\frac{1}{3})^{i-1}(-\displaystyle\frac{1}{3})}$

    The doubt I've got is about the "negative" part of the summation, I know how to solve the geometric progression, but I'm not sure about the other part. I've proceeded this way:
    $\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=1-2+7-20$
    This is what I did: $\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=s_{-1}+s_{-2}+s_{-3}+s_{0}$

    I'm not sure about this, may I should do it this other way?:
    $\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=a_{0}+a_{-1}+a_{-2}+a_{-3}$
    I mean, taking the values of the sequence for the series for every "i", and making that summation but the accumulation? cause I think it has not much sense what I did before, I mean there is no accumulation for the negative part, right?

    Bye there, and thanks for your help.
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  2. #2
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    $\displaystyle \displaystyle \sum_{i=-3}^{\infty}\left(\frac{-1}{3}\right)^i = \sum_{i=0}^{\infty}\left(\frac{-1}{3}\right)^{i-3}=$

    $\displaystyle \displaystyle \left(\frac{-1}{3}\right)^{-3}\sum_{i=0}^{\infty}\left(\frac{-1}{3}\right)^{i}=\frac{a}{1-r}$

    $\displaystyle \displaystyle a=\left(\frac{-1}{3}\right)^{-3}, \ r=\frac{-1}{3}$

    Geometric series.
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  3. #3
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    $\displaystyle \displaystyle \sum_{k \ge {-3}}\left(-\frac{1}{3}\right)^k = -21+\sum_{k \ge 0}\left(-\frac{1}{3}\right)^k = -21+\sum_{k \ge 0}x^k\bigg|_{x = -\frac{1}{3}} = -21+\frac{1}{1-x}\bigg|_{x = -\frac{1}{3}} = -\frac{81}{4}.$
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  4. #4
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    This is the case:
    If $\displaystyle J\in\mathbb{Z}~\&~|r|<1$ then $\displaystyle \displaystyle
    \sum\limits_{n = J}^\infty {ar^n } = \frac{{ar^J }}{{1 - r}}$.

    It is that easy.
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  5. #5
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    Also...

    $\displaystyle \displaystyle\left[\frac{-1}{3}\right]^i=\left[-3\right]^{-i}$

    $\displaystyle [-3]^3+[-3]^2+....$

    is a geometric series with first term $\displaystyle [-3]^3$ and common ratio $\displaystyle [-3]^{-1}$

    $\displaystyle \displaystyle\ S_{\infty}=\frac{[-3]^3}{1-[-3]^{-1}}$
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  6. #6
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    This is a popular thread!

    Quote Originally Posted by Ulysses View Post
    Hi there. I have some doubts with this exercise. I have to make the summation for:

    $\displaystyle \displaystyle\sum_{i=-3}^\infty{\left(-\displaystyle\frac{1}{3}\right)^i}$

    ...

    Bye there, and thanks for your help.

    A solution for those who can't remember the results or requirements for a geometric series.

    Let $\displaystyle \displaystyle S =\sum_{k=-3}^\infty\left({-{{1}\over{3}}\right)^k}.\quad\quad\quad\quad\dots (1)$

    Remember, $\displaystyle \displaystyle S$ is equal to the sum we are to evaluate.

    Multiply both sides of equation (1) by $\displaystyle \displaystyle \left({-{{1}\over{3}}\right)^{-1}=-3$ .

    $\displaystyle \displaystyle -3S =\left({-{{1}\over{3}}\right)^{-1}}\sum_{k=-3}^\infty\left({-{{1}\over{3}}\right)^{k}}$


    $\displaystyle \displaystyle -3S =\sum_{k=-3}^\infty\left({-{{1}\over{3}}\right)^{k-1}\quad\quad\quad\quad\dots (2)$

    The series in equation 2 actually starts at $\displaystyle \displaystyle \left({-{{1}\over{3}}\right)^{-4}$, so let's change the index to reflect that.

    $\displaystyle \displaystyle -3S =\sum_{j=-4}^\infty\left({-{{1}\over{3}}\right)^{j}$

    Separate the first term, $\displaystyle \displaystyle \left({-{{1}\over{3}}\right)^{-4}$, of this series out from the summation.

    $\displaystyle \displaystyle -3S =\left({-{{1}\over{3}}\right)^{-4}+\sum_{j=-3}^\infty\left({-{{1}\over{3}}\right)^{j}$

    The above sum is $\displaystyle \displaystyle S$, and $\displaystyle \displaystyle \left({-{{1}\over{3}}\right)^{-4}=81,$ so we have:

    $\displaystyle \displaystyle -3S =81+S.$

    Solving this for $\displaystyle \displaystyle S$ gives $\displaystyle \displaystyle S=-{{81}\over{4}}\,.$


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  7. #7
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    Quote Originally Posted by Ulysses View Post
    Hi there. I have some doubts with this exercise. I have to make the summation for:

    $\displaystyle \displaystyle\sum_{i=-3}^\infty{(-\displaystyle\frac{1}{3})^i}$

    I've solved this way:

    $\displaystyle \displaystyle\sum_{i=-3}^{\infty}{(-\displaystyle\frac{1}{3})^i}=\displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}+\displaystyle\sum_{i= 1}^{\infty}{(-\displaystyle\frac{1}{3})^{i-1}(-\displaystyle\frac{1}{3})}$

    The doubt I've got is about the "negative" part of the summation, I know how to solve the geometric progression, but I'm not sure about the other part. I've proceeded this way:
    $\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=1-2+7-20$
    This is what I did: $\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=s_{-1}+s_{-2}+s_{-3}+s_{0}$
    Your formula is correct but that does not give what you have above. When i= 0, $\displaystyle \left(-\frac{1}{3}\right)^0= 1$, of course, but when i= -1, $\displaystyle \left(-\frac{1}{3}\right)^{-1}= -3$, when i= -2, $\displaystyle \left(-\frac{1}{3}\right)^{-2}= 9$, and when i= -3, $\displaystyle \left(-\frac{1}{3}\right)^{-3}= -27$. I have no idea where you got "2", "7", and "20".
    That sum is 1- 3+ 9- 27= -20

    I'm not sure about this, may I should do it this other way?:
    $\displaystyle \displaystyle\sum_{i=-3}^0{(-\displaystyle\frac{1}{3})^i}=a_{0}+a_{-1}+a_{-2}+a_{-3}$
    I mean, taking the values of the sequence for the series for every "i", and making that summation but the accumulation? cause I think it has not much sense what I did before, I mean there is no accumulation for the negative part, right?

    Bye there, and thanks for your help.
    Follow Math Help Forum on Facebook and Google+

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