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    I'm having some trouble with the summation of  \displaystyle\sum_{n=1}^{N} \dfrac{1}{x^{n+1}}
    I have that

     \displaystyle\sum_{n=1}^{N} \dfrac{1}{n^2} < \dfrac{e}{2} \displaystyle\sum_{n=1}^{N} \displaystyle\int_{e^N}^{\infty} \dfrac{\ln{x}}{x^{n+1}} \mathrm{d} x

    How ever I need to manipulate this into what we see in the question (STEP II 2003 Q7, the last part). I'm having some trouble, since

     \dfrac{e}{2} \displaystyle\sum_{n=1}^{N} \displaystyle\int_{e^N}^{\infty} \dfrac{\ln{x}}{x^{n+1}} \mathrm{d} x = \dfrac{e}{2} \displaystyle\int_{e^N}^{\infty} \ln{x} \displaystyle\sum_{n=1}^{N} \dfrac{1}{x^{n+1}} \mathrm{d} x

    But I don't know how to show that

     \displaystyle\sum_{n=1}^{N} \dfrac{1}{x^{n+1}} = \dfrac{1 - x^{-N}}{x^2 - x}

    I think yesterday (when I was doing this Q) I found that

     \dfrac{1 - x^{-N}}{x^2 - x} = x^{-2} + x^{-3} + ... + x^{-(N+1)}

    By using long devision however I had a hard time approaching it from original sum.
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    (Original post by Louisb19)

    But I don't know how to show that

     \displaystyle\sum_{n=1}^{N} \dfrac{1}{x^{n+1}} = \dfrac{1 - x^{-N}}{x^2 - x}
    What is

     \displaystyle\sum_{n=1}^{N} u^{n+1} ?

    Now substitute u = 1/x
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    (Original post by Louisb19)
    I'm having some trouble with the summation of  \displaystyle\sum_{n=1}^{N} \dfrac{1}{x^{n+1}}
    I have that

     \displaystyle\sum_{n=1}^{N} \dfrac{1}{n^2} < \dfrac{e}{2} \displaystyle\sum_{n=1}^{N} \displaystyle\int_{e^N}^{\infty} \dfrac{\ln{x}}{x^{n+1}} \mathrm{d} x

    How ever I need to manipulate this into what we see in the question (STEP II 2003 Q7, the last part). I'm having some trouble, since

     \dfrac{e}{2} \displaystyle\sum_{n=1}^{N} \displaystyle\int_{e^N}^{\infty} \dfrac{\ln{x}}{x^{n+1}} \mathrm{d} x = \dfrac{e}{2} \displaystyle\int_{e^N}^{\infty} \ln{x} \displaystyle\sum_{n=1}^{N} \dfrac{1}{x^{n+1}} \mathrm{d} x

    But I don't know how to show that

     \displaystyle\sum_{n=1}^{N} \dfrac{1}{x^{n+1}} = \dfrac{1 - x^{-N}}{x^2 - x}

    I think yesterday (when I was doing this Q) I found that

     \dfrac{1 - x^{-N}}{x^2 - x} = x^{-2} + x^{-3} + ... + x^{-(N+1)}

    By using long devision however I had a hard time approaching it from original sum.
    Gp


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