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Math Help - Summation help (urgent)

  1. #1
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    Summation help (urgent)

    can someone help me go through this question i cant figure it out, i need help asap

     \sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}
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    Quote Originally Posted by intomath View Post
    can someone help me go through this question i cant figure it out, i need help asap

     \sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}
    What are you trying to do with it? Show it converges or sum it?
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    all it says in my paper is Find
    <br /> <br />
\sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}<br />
    It carries 8 marks though, maybe you have to show both?
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    Quote Originally Posted by intomath View Post
    all it says in my paper is Find
    <br /> <br />
\sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}<br />
    It carries 8 marks though, maybe you have to show both?
    Well obviously if you evaluate the sum, it's convergent... So yes, you'd show both.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Prove It View Post
    Well obviously if you evaluate the sum, it's convergent... So yes, you'd show both.
    \sum_{n=1}^{\infty}\frac{1}{3^{n-2}n}\leq\sum_{n=1}^{\infty}\frac{1}{3^{n-2}}=\frac{9}{2}

    Now for the sum rewrite it as

    9\sum_{n=1}^{\infty}\frac{\left(\frac{1}{3}\right)  ^n}{n}

    Now it is known that

    \sum_{n=1}^{\infty}\frac{\left(x\in[-1,1)\right)^n}{n}=-\ln(1-x)

    So our sum is

    9\ln\left(\frac{3}{2}\right)
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  6. #6
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    Quote Originally Posted by intomath View Post
    all it says in my paper is Find
    <br /> <br />
\sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}<br />
    It carries 8 marks though, maybe you have to show both?
    Finding the value of the sum will answer the question, which is all that you're required to do.
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  7. #7
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    Quote Originally Posted by intomath View Post
    can someone help me go through this question i cant figure it out, i need help asap

     \sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}
    Sum of infinite geometric series:

    \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} for |r| < 1.

    Integrate both sides (there are some technical issues regarding the justification of changing the order of operations):

    \sum_{n=0}^{\infty} \frac{r^{n+1}}{n+1} = - \ln (1 - r) + C for |r| < 1

    Substitute r = 0: ~ 0 = 0 + C \Rightarrow C = 0

    \Rightarrow \sum_{n={\color{red}1}}^{\infty} \frac{r^n}{n} = - \ln (1 - r) for |r| < 1.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Sum of infinite geometric series:

    \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} for |r| < 1.

    Integrate both sides (there are some technical issues regarding the justification of changing the order of operations):

    \sum_{n=0}^{\infty} \frac{r^{n+1}}{n+1} = - \ln (1 - r) + C for |r| < 1

    Substitute r = 0: ~ 0 = 0 + C \Rightarrow C = 0

    \Rightarrow \sum_{n={\color{red}1}}^{\infty} \frac{r^n}{n} = - \ln (1 - r) for |r| < 1.
    -\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}~|x|<1\cup\left  \{-1\right\}

    ...just saying
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