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Math Help - Estimating the sum of a series

  1. #1
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    Estimating the sum of a series

    Hello once again,

    I have this problem that wants me to estimate the sum of a given series within 0.0001.

    "Consider the series \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}. Estimate the sum of the series within 0.0001."

    I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.



    Thanks.
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  2. #2
    Moo
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    Hello !

    Quote Originally Posted by auslmar View Post
    Hello once again,

    I have this problem that wants me to estimate the sum of a given series within 0.0001.

    "Consider the series \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}. Estimate the sum of the series within 0.0001."

    I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.



    Thanks.
    The sum is equal to :

    4 \cdot \sum_{k=1}^{\infty} \frac{(-1)^k}{3^k \cdot k!}=4 \cdot \sum_{k={\color{red}1}}^{\infty} \frac{\left(-\frac 13\right)^k}{k!}.

    Now, I will use a known series :

    e^x=\sum_{k={\color{red}0}}^{\infty} \frac{x^k}{k!}=1+\sum_{k={\color{red}1}}^{\infty} \frac{x^k}{k!}

    \implies \sum_{k={\color{red}1}}^{\infty} \frac{x^k}{k!}=e^x-1

    Here, x=-\tfrac 13. Can you finish it ?
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  3. #3
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    Quote Originally Posted by auslmar View Post
    Hello once again,

    I have this problem that wants me to estimate the sum of a given series within 0.0001.

    "Consider the series \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{(3^k)*k!}. Estimate the sum of the series within 0.0001."

    I can't find any examples in my textbook that seem to explain how I would go about this. Can anyone explain how to approach this problem? I would greatly appreciate any help.



    Thanks.
    This is an alternating series of decreasing terms, so the error in a partial sum S_n of the first n terms is less than the first neglected term.

    So:

    |S-S_n|<\frac{4}{3^{n+1} (n+1)!}

    So we want:

    |S-S_n|<0.0001

    we find the smallest integer n such that:

    \frac{4}{3^{n+1} (n+1)!}\le 0.0001

    which by trial and error we find is n=?

    RonL
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  4. #4
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    Hello, auslmar!

    An elaboration of moo's excellent solution . . .


    Consider the series: . \sum_{k=1}^{\infty}\frac{{4(-1)^k}}{3^k\cdot k!}

    Estimate the sum of the series within 0.0001.

    We have: . S \;=\;-\frac{4}{3\cdot1!} + \frac{4}{3^2\cdot2!} - \frac{4}{3^3\cdot3!} + \frac{4}{3^4\cdot4!} - \hdots

    . . . . . . . . S \;=\;4\underbrace{\left[-\frac{\frac{1}{3}}{1!} + \frac{\frac{1}{3^2}}{2!} - \frac{\frac{1}{3^3}}{3!} + \frac{\frac{1}{3^4}}{4!}\hdots\right]}_{\text{Infinite series of: }e^{\text{-}\frac{1}{3}}-1}


    Therefore: . S\;=\;4\left(e^{-\frac{1}{3}} - 1\right) \;\approx\;-1.1339

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  5. #5
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    Sorry for the late response, I had to hit the hay.

    Thanks to everyone for the help. I'm not sure I actually understand Moo's or Soroban's examples, but CaptainBlack's makes the most sense to me. If I understand correctly, the goal is to determine the sufficient nth term of the series that would be less than or equal to within 0.0001 of the "actual" sum. Is this notion correct? I applied it to a similar problem with correct results.

    Thanks again!

    Quote Originally Posted by CaptainBlack View Post
    This is an alternating series of decreasing terms, so the error in a partial sum S_n of the first n terms is less than the first neglected term.

    So:

    |S-S_n|<\frac{4}{3^{n+1} (n+1)!}

    So we want:

    |S-S_n|<0.0001

    we find the smallest integer n such that:

    \frac{4}{3^{n+1} (n+1)!}\le 0.0001

    which by trial and error we find is n=?

    RonL
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