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Thread: Sum of series .. keep getting wrong answer

  1. #1
    Ted
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    Sum of series .. keep getting wrong answer

    Hello,
    Question :
    Find the sum of the convergent series ( (1/e)^n + ( 1 / n(n+1) ) ) from n=1 to n=infinity
    Clearly, the series is a sum of two convergent series
    First series is geometric series with sum = a / (1-r) = 1 / ( 1 - (1/e) ) = e / (e-1)
    Second series is telescoping series with Sn = 1 - ( 1 / (n+1) ) which approach 1 as n approach infinity
    So the sum is the sum of the two series 1 + ( e/(e-1) )
    But the book and wolframalpha keep saying that the sum is e/(e-1) !
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  2. #2
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    Re: Sum of series .. keep getting wrong answer

    Quote Originally Posted by Ted View Post
    Hello,
    Question :
    Find the sum of the convergent series ( (1/e)^n + ( 1 / n(n+1) ) ) from n=1 to n=infinity
    Clearly, the series is a sum of two convergent series
    First series is geometric series with sum = a / (1-r) = 1 / ( 1 - (1/e) ) = e / (e-1)
    Second series is telescoping series with Sn = 1 - ( 1 / (n+1) ) which approach 1 as n approach infinity
    So the sum is the sum of the two series 1 + ( e/(e-1) )
    But the book and wolframalpha keep saying that the sum is e/(e-1) !
    Because the telescoping sequence isn't defined for n=0, you are summing from 1 to infinity. Not 0 to infinity.

    $\displaystyle{\sum_{n=1}^{\infty}}\left(\dfrac 1 e\right)^n = \dfrac 1 {e-1}$

    $1 + \dfrac 1 {e-1} = \dfrac e {e-1}$
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  3. #3
    Ted
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    Re: Sum of series .. keep getting wrong answer

    Quote Originally Posted by romsek View Post
    Because the telescoping sequence isn't defined for n=0, you are summing from 1 to infinity. Not 0 to infinity.
    $\displaystyle{\sum_{n=1}^{\infty}}\left(\dfrac 1 e\right)^n = \dfrac 1 {e-1}$
    I know that why does it start from n=1
    How did you calculate that sum ?
    Applying geometric series formula gives e / (e-1)
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    Re: Sum of series .. keep getting wrong answer

    Quote Originally Posted by Ted View Post
    I know that why does it start from n=1
    How did you calculate that sum ?
    Applying geometric series formula gives e / (e-1)
    it gets $\dfrac e {e-1}$ when you start from $n=0$

    it gets $\dfrac 1 {e-1}$ when you start from $n=1$
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  5. #5
    Ted
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    Re: Sum of series .. keep getting wrong answer

    Why is that?
    My professor did not mention that
    just give us the form of geometric series a * (r)^n and the formula of the sum a / (1-r)
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    Re: Sum of series .. keep getting wrong answer

    Quote Originally Posted by Ted View Post
    Why is that?
    My professor did not mention that
    just give us the form of geometric series a * (r)^n and the formula of the sum a / (1-r)
    you can wiki all this faster than I can type it out and have you argue about it with me.
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    Re: Sum of series .. keep getting wrong answer

    $\displaystyle \sum_{n=1}^\infty r^n = \left(\sum_{n=0}^\infty r^n\right) - r^0 = \left(\sum_{n=0}^\infty r^n\right)-1 = \dfrac{1}{1-r}-1$

    In your case, $\displaystyle r = \dfrac{1}{e}$, so you have:

    $\displaystyle \sum_{n=1}^\infty \left(\dfrac{1}{e}\right)^n = \dfrac{1}{1-\tfrac{1}{e}} - 1 = \dfrac{e}{e-1} - 1 = \dfrac{1}{e-1}$

    Just as romsek said.
    Thanks from Ted
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  8. #8
    Ted
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    Re: Sum of series .. keep getting wrong answer

    All what do you need to say is that the formula apply when n starts with 0 and i will get it without any arguing.
    Thank you all
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