Generalize $sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}$












1












$begingroup$


I was looking at this wolfram site on section [25]



A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)



where $kge1$



$$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$



I was trying to generalize $(1)$



Let generalize $(1)$,



where $xge0$



$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$



and hoping to find a closed form but I could only got a partial of it



$k=1$



$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$



How do we go about to find the closed form for $(2)$?



I am assuming $(2)$, may take the closed form of



$$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$






sequences-and-series







share|cite|improve this question











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    1












    $begingroup$


    I was looking at this wolfram site on section [25]



    A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)



    where $kge1$



    $$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$



    I was trying to generalize $(1)$



    Let generalize $(1)$,



    where $xge0$



    $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$



    and hoping to find a closed form but I could only got a partial of it



    $k=1$



    $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$



    How do we go about to find the closed form for $(2)$?



    I am assuming $(2)$, may take the closed form of



    $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$






    sequences-and-series







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I was looking at this wolfram site on section [25]



      A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)



      where $kge1$



      $$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$



      I was trying to generalize $(1)$



      Let generalize $(1)$,



      where $xge0$



      $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$



      and hoping to find a closed form but I could only got a partial of it



      $k=1$



      $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$



      How do we go about to find the closed form for $(2)$?



      I am assuming $(2)$, may take the closed form of



      $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$






      sequences-and-series







      share|cite|improve this question











      $endgroup$




      I was looking at this wolfram site on section [25]



      A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)



      where $kge1$



      $$sum_{n=1}^{infty}frac{H_n}{(n+1)(n+2)cdots(n+k+1)}=frac{1}{k!k^2}tag1$$



      I was trying to generalize $(1)$



      Let generalize $(1)$,



      where $xge0$



      $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}tag2$$



      and hoping to find a closed form but I could only got a partial of it



      $k=1$



      $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)}=-frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2tag3$$



      How do we go about to find the closed form for $(2)$?



      I am assuming $(2)$, may take the closed form of



      $$sum_{n=1}^{infty}frac{H_n}{(n+1+x)(n+2+x)cdots(n+k+1+x)}=G(x)-sum_{j=0}^{k}(-1)^j{k choose j}(H_{x+j})^2tag4$$






      sequences-and-series




      sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 2 at 20:53







      user583851

















      asked Jan 2 at 20:33









      user583851user583851

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      508110






















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          $begingroup$

          By summation by parts the computation of $(2)$ boils down to the computation of
          $$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
          or
          $$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
          or
          $$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
          which only depends on the Beta function and its partial derivatives. In particular the last expression equals



          $$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$






          share|cite|improve this answer









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            $begingroup$

            By summation by parts the computation of $(2)$ boils down to the computation of
            $$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
            or
            $$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
            or
            $$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
            which only depends on the Beta function and its partial derivatives. In particular the last expression equals



            $$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$






            share|cite|improve this answer









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              1












              $begingroup$

              By summation by parts the computation of $(2)$ boils down to the computation of
              $$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
              or
              $$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
              or
              $$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
              which only depends on the Beta function and its partial derivatives. In particular the last expression equals



              $$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                By summation by parts the computation of $(2)$ boils down to the computation of
                $$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
                or
                $$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
                or
                $$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
                which only depends on the Beta function and its partial derivatives. In particular the last expression equals



                $$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$






                share|cite|improve this answer









                $endgroup$



                By summation by parts the computation of $(2)$ boils down to the computation of
                $$ sum_{ngeq 1}frac{1}{(n+1)(n+1+x)(n+2+x)cdots(n+k+x)}=sum_{ngeq 1}frac{Gamma(n+1+x)}{(n+1)Gamma(n+k+x)} $$
                or
                $$ frac{1}{Gamma(k-1)}sum_{ngeq 1}frac{B(n+1+x,k-1)}{n+1}=frac{1}{Gamma(k-1)}int_{0}^{1}sum_{ngeq 1}frac{(1-z)^{k-2}z^{n+x+2}}{n+1},dz $$
                or
                $$ frac{1}{Gamma(k-1)}int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-log(1-z)),dz $$
                which only depends on the Beta function and its partial derivatives. In particular the last expression equals



                $$ -frac{Gamma(x+2)Gamma(k-1)}{Gamma(x+k+2)}left[2+x+(x+k+1)(H_{k-2}-H_{k+x})right]. $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 21:54









                Jack D'AurizioJack D'Aurizio

                291k33284666




                291k33284666






























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