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Math Help - Series...

  1. #1
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    Series...

    Determine whether the series converges or diverges. If the series converges, find its limit. Justify your answers.

     \sum_{j=2}^{\infty} \frac{3^j}{4^{j+1}}

    So first I did this, although I'm not sure if I'm going in the right direction, although I believe I am...

    =  \frac {3^2}{4^3} (1 + \frac {3}{4} + \frac {3^2}{4^2} + \frac {3^3}{4^3} ...) then I can't figure out what the stuff inside the parenthesis equals out to be, then I could figure out the answer.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by larson View Post
    Determine whether the series converges or diverges. If the series converges, find its limit. Justify your answers.

     \sum_{j=2}^{\infty} \frac{3^j}{4^{j+1}}

    So first I did this, although I'm not sure if I'm going in the right direction, although I believe I am...

    =  \frac {3^2}{4^3} (1 + \frac {3}{4} + \frac {3^2}{4^2} + \frac {3^3}{4^3} ...) then I can't figure out what the stuff inside the parenthesis equals out to be, then I could figure out the answer.
    Here I will give you big geometric hint...this can be rewritten as \sum_{j=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3}{  4}\bigg)^{j}
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    Quote Originally Posted by Mathstud28 View Post
    Here I will give you big geometric hint...this can be rewritten as \sum_{j=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3}{  4}\bigg)^{j}
    is what I wrote correct though? Even if what you wrote is geometric, where does that put me?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by larson View Post
    is what I wrote correct though? Even if what you wrote is geometric, where does that put me?
    its convergent since \bigg|\frac{3}{4}\bigg|<1 which is the definition of the convergent geometric series
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by larson View Post
    is what I wrote correct though? Even if what you wrote is geometric, where does that put me?
    \sum_{n=0}^{\infty}r^n=\frac{1}{1-r} if r < 1
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    its convergent since \bigg|\frac{3}{4}\bigg|<1 which is the definition of the convergent geometric series
    and the sum would be \bigg(\frac{3}{4}\bigg)^2\cdot\bigg(\frac{1}{4}\bi  gg)\cdot\frac{1}{1-\frac{3}{4}}
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    Quote Originally Posted by Mathstud28 View Post
    and the sum would be \bigg(\frac{3}{4}\bigg)^2\cdot\bigg(\frac{1}{4}\bi  gg)\cdot\frac{1}{1-\frac{3}{4}}
    I also just figured it out the way I was first doing it by figuring out how to do series / sums on my TI-83 Plus . I figured out what was the in the parenthesis was 4, so  \frac {(4)3^2}{4^2} = \frac {9}{16}.
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    Quote Originally Posted by Mathstud28 View Post
    Here I will give you big geometric hint...this can be rewritten as \sum_{j=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3}{  4}\bigg)^{j}
    Could I ask you real fast how you knew to find that  \frac {1}{4}
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by larson View Post
    Could I ask you real fast how you knew to find that  \frac {1}{4}
    4^{j + 1} = 4 \cdot 4^j. we wanted the 4^j, so we pulled the 4 out from the denominator making the \frac 14
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by larson View Post
    Could I ask you real fast how you knew to find that  \frac {1}{4}
    sure you have \sum_{n=2}^{\infty}\frac{3^n}{4^{n+1}}=\sum_{n=2}^  {\infty}\frac{3^n}{4\cdot{4^{n}}}=\sum_{n=2}^{\inf  ty}\frac{1}{4}\cdot\frac{3^n}{4^n} =\sum_{n=2}^{\infty}\frac{1}{4}\cdot\bigg(\frac{3}  {4}\bigg)^n
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  11. #11
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    You may modify the sum by changing the lower limit to 0. From there you'll get a direct geometric series.
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  12. #12
    Moo
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    Quote Originally Posted by Krizalid View Post
    You may modify the sum by changing the lower limit to 0. From there you'll get a direct geometric series.
    You don't need to, you just have to say that :

    \sum_{k=m}^n x^k=x^m \frac{1-x^{m+n+1}}{1-x}

    Oh...I'm not sure I've understood correctly the thing actually
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    You don't need to, you just have to say that :

    \sum_{k=m}^n x^k=x^m \frac{1-x^{m+n+1}}{1-x}

    Oh...I'm not sure I've understood correctly the thing actually
    Actualy...I am pretty sure \sum_{m=p}^{\infty}x^{n}=x^{p}\frac{1}{1-x},\forall{|x|}<1
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  14. #14
    Moo
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    Of course...

    I was just giving the general formula...
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Of course...

    I was just giving the general formula...
    OK!
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