$sum_{n = 1}^{D - 1} frac{n}{D - n}$ written as a function of $D$? [closed]












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$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$



Is it possible to reduce this summation to a function of $D$?










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closed as unclear what you're asking by amWhy, Saad, Alexander Gruber Nov 30 '18 at 3:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
    – Alex Vong
    Nov 28 '18 at 12:22
















0














$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$



Is it possible to reduce this summation to a function of $D$?










share|cite|improve this question















closed as unclear what you're asking by amWhy, Saad, Alexander Gruber Nov 30 '18 at 3:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 3




    Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
    – Alex Vong
    Nov 28 '18 at 12:22














0












0








0







$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$



Is it possible to reduce this summation to a function of $D$?










share|cite|improve this question















$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$



Is it possible to reduce this summation to a function of $D$?







summation






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edited Nov 28 '18 at 12:20









amWhy

192k28225439




192k28225439










asked Nov 28 '18 at 12:05









bestscammer5

1




1




closed as unclear what you're asking by amWhy, Saad, Alexander Gruber Nov 30 '18 at 3:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as unclear what you're asking by amWhy, Saad, Alexander Gruber Nov 30 '18 at 3:36


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 3




    Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
    – Alex Vong
    Nov 28 '18 at 12:22














  • 3




    Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
    – Alex Vong
    Nov 28 '18 at 12:22








3




3




Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
– Alex Vong
Nov 28 '18 at 12:22




Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
– Alex Vong
Nov 28 '18 at 12:22










1 Answer
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$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$

where $H_n$ is the $n$th harmonic number.






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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

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    $$sum_{n = 1}^{D - 1} frac{n}{D - n}\
    =sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
    =-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
    =-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
    = -D+1 + D H_{D-1}$$

    where $H_n$ is the $n$th harmonic number.






    share|cite|improve this answer


























      1














      $$sum_{n = 1}^{D - 1} frac{n}{D - n}\
      =sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
      =-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
      =-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
      = -D+1 + D H_{D-1}$$

      where $H_n$ is the $n$th harmonic number.






      share|cite|improve this answer
























        1












        1








        1






        $$sum_{n = 1}^{D - 1} frac{n}{D - n}\
        =sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
        =-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
        =-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
        = -D+1 + D H_{D-1}$$

        where $H_n$ is the $n$th harmonic number.






        share|cite|improve this answer












        $$sum_{n = 1}^{D - 1} frac{n}{D - n}\
        =sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
        =-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
        =-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
        = -D+1 + D H_{D-1}$$

        where $H_n$ is the $n$th harmonic number.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 12:33









        Calvin Khor

        11.2k21438




        11.2k21438















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