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Math Help - I want to see if I'm right...

  1. #1
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    I want to see if I'm right...

    I am given the series (e^n)/(3^(n-1))...Don't forget the Riemann sum symbol in front of this. It appears to be a geometric series. This would mean its ar^n
    and a = 1/3 and r = e. I then do a bit of manipulation and get 1/3-1/3(e^n) and then Sn = (1-e^n)/(3-3e)

    since r>1, the sequence {r^n} is divergent. Lim n--> infinity Sn does not exist. The geometric series is divergent.
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by Undefdisfigure View Post
    I am given the series (e^n)/(3^(n-1))...Don't forget the Riemann sum symbol in front of this. It appears to be a geometric series. This would mean its ar^n
    and a = 1/3 and r = e. I then do a bit of manipulation and get 1/3-1/3(e^n) and then Sn = (1-e^n)/(3-3e)

    since r>1, the sequence {r^n} is divergent. Lim n--> infinity Sn does not exist. The geometric series is divergent.
    \sum_{n=0}^{\infty}3(\frac{e}{3})^n = \frac{3}{1-\frac{e}{3}} \Rightarrow \frac{9}{3-e}

    It is a geometric series - how is it divergent?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Undefdisfigure View Post
    I am given the series (e^n)/(3^(n-1))...Don't forget the Riemann sum symbol in front of this. It appears to be a geometric series. This would mean its ar^n
    and a = 1/3 and r = e. I then do a bit of manipulation and get 1/3-1/3(e^n) and then Sn = (1-e^n)/(3-3e)

    since r>1, the sequence {r^n} is divergent. Lim n--> infinity Sn does not exist. The geometric series is divergent.
    Your mistake appears to be in taking the limit of your sum formula:
    \sum_{i = 0}^n \frac{e^i}{3^{i - 1}}

    Notice that your r is not e, but e/3 < 1. Thus when we take
    \lim_{n \to \infty} \sum_{i = 0}^n \frac{e^i}{3^{i - 1}} = \lim_{n \to \infty} \sum_{n = 0}^n 3 \left ( \frac{e}{3} \right )^i

    = \lim_{n \to \infty} \frac{3 \left (1 - \left [ \frac{e}{3} \right ] ^{n + 1} \right )}{1 - \frac{e}{3}}

    = \frac{3}{1 - \frac{e}{3}}
    (the e/3 term in the numerator vanishes because e/3 < 1)

    = \frac{9}{3 - e}

    -Dan
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  4. #4
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    I have another question. I have to find a telescoping sum of ln n/(n+1). I do not see how to do a partial fraction decomposition on this. Do I have to take the derivative of this expression? I have to find if this expression is convergent and if it is, to find its sum.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Undefdisfigure View Post
    I have another question. I have to find a telescoping sum of ln n/(n+1). I do not see how to do a partial fraction decomposition on this. Do I have to take the derivative of this expression? I have to find if this expression is convergent and if it is, to find its sum.
    what partial fraction? recall, \log_a \left( \frac xy \right) = \log_a x - \log_a y

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