Prove that $sum^n_{k=0}frac{(-1)^k}{k+x}binom{n}{k}=frac{n!}{x(x+1)cdots(x+n)}$.











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Given the following formula
$$
sum^n_{k=0}frac{(-1)^k}{k+x}binom{n}{k},.
$$

How can I show that this is equal to
$$
frac{n!}{x(x+1)cdots(x+n)},?
$$










share|cite|improve this question




















  • 1




    Both functions have the same simple poles, with the same residues. Or you may just use induction on $n$.
    – Jack D'Aurizio
    Nov 19 at 16:10












  • One of several links that you may consult is at this MSE post.
    – Marko Riedel
    Nov 19 at 16:39










  • Or alternatively, this MSE post II.
    – Marko Riedel
    Nov 19 at 16:47















up vote
4
down vote

favorite
1












Given the following formula
$$
sum^n_{k=0}frac{(-1)^k}{k+x}binom{n}{k},.
$$

How can I show that this is equal to
$$
frac{n!}{x(x+1)cdots(x+n)},?
$$










share|cite|improve this question




















  • 1




    Both functions have the same simple poles, with the same residues. Or you may just use induction on $n$.
    – Jack D'Aurizio
    Nov 19 at 16:10












  • One of several links that you may consult is at this MSE post.
    – Marko Riedel
    Nov 19 at 16:39










  • Or alternatively, this MSE post II.
    – Marko Riedel
    Nov 19 at 16:47













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





Given the following formula
$$
sum^n_{k=0}frac{(-1)^k}{k+x}binom{n}{k},.
$$

How can I show that this is equal to
$$
frac{n!}{x(x+1)cdots(x+n)},?
$$










share|cite|improve this question















Given the following formula
$$
sum^n_{k=0}frac{(-1)^k}{k+x}binom{n}{k},.
$$

How can I show that this is equal to
$$
frac{n!}{x(x+1)cdots(x+n)},?
$$







summation factorial partial-fractions rational-functions






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edited Nov 22 at 17:12









Batominovski

33.5k33292




33.5k33292










asked Nov 19 at 15:51









RedPen

212112




212112








  • 1




    Both functions have the same simple poles, with the same residues. Or you may just use induction on $n$.
    – Jack D'Aurizio
    Nov 19 at 16:10












  • One of several links that you may consult is at this MSE post.
    – Marko Riedel
    Nov 19 at 16:39










  • Or alternatively, this MSE post II.
    – Marko Riedel
    Nov 19 at 16:47














  • 1




    Both functions have the same simple poles, with the same residues. Or you may just use induction on $n$.
    – Jack D'Aurizio
    Nov 19 at 16:10












  • One of several links that you may consult is at this MSE post.
    – Marko Riedel
    Nov 19 at 16:39










  • Or alternatively, this MSE post II.
    – Marko Riedel
    Nov 19 at 16:47








1




1




Both functions have the same simple poles, with the same residues. Or you may just use induction on $n$.
– Jack D'Aurizio
Nov 19 at 16:10






Both functions have the same simple poles, with the same residues. Or you may just use induction on $n$.
– Jack D'Aurizio
Nov 19 at 16:10














One of several links that you may consult is at this MSE post.
– Marko Riedel
Nov 19 at 16:39




One of several links that you may consult is at this MSE post.
– Marko Riedel
Nov 19 at 16:39












Or alternatively, this MSE post II.
– Marko Riedel
Nov 19 at 16:47




Or alternatively, this MSE post II.
– Marko Riedel
Nov 19 at 16:47










3 Answers
3






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3
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accepted










Induction step:



$$begin{align}
sum_{k=0}^{n+1}&frac{(-1)^k}{x+k}binom{n+1}k=frac1x+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}left[binom nk+binom n{k-1}right]
\&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}binom{n}{k-1}
\&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-sum_{k=0}^{n-1}frac{(-1)^k}{(x+1)+k}binom{n}{k}
\&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-frac{n!}{(x+1)(x+2)cdots(x+n+1)}+frac{(-1)^n}{x+n+1}
\&=frac{n!(x+n+1)-n!x}{x(x+1)cdots(x+n+1)}=frac{(n+1)!}{x(x+1)cdots(x+n+1)}
end{align}$$






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    down vote













    Consider the (unique) polynomial $p(x)inmathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,ldots,n$. Clearly, $p(x)$ is the constant polynomial $1$.



    However, using Lagrange interpolation, we have
    $$p(x)=sum_{k=0}^n,p(-k),frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)},,$$
    where $[n]:={0,1,2,ldots,n}$. This means
    $$1=sum_{k=0}^n,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)}=sum_{k=0}^n,(-1)^k,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{k!,(n-k)!},.$$
    Multiplying both sides by $dfrac{n!}{prodlimits_{jin[n]},(x+j)}$ yields
    $$frac{n!}{prodlimits_{j=0}^n,(x+j)}=sum_{k=0}^n,(-1)^k,left(frac{n!}{k!,(n-k)!}right),frac{1}{x+k}=sum_{k=0}^n,binom{n}{k},frac{(-1)^k}{x+k},.$$






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      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{ic}{mathrm{i}}
      newcommand{mc}[1]{mathcal{#1}}
      newcommand{mrm}[1]{mathrm{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$




      With $ds{Repars{x} > 0}$:




      begin{align}
      &bbox[10px,#ffd]{sum_{k = 0}^{n}{pars{-1}^{k} over k + x}{n choose k}} =
      sum_{k = 0}^{n}pars{-1}^{k}
      pars{int_{0}^{1}t^{k + x - 1},dd t}{n choose k}
      \[5mm] = &
      int_{0}^{1}t^{x - 1}sum_{k = 0}^{n}
      {n choose k}pars{-t}^{k},dd t
      \[5mm] = &
      int_{0}^{1}t^{x - 1},pars{1 - t}^{n},dd t =
      mrm{B}pars{x,n + 1} pars{~mrm{B}: Beta Function~}
      \[5mm] = &
      {Gammapars{x}Gammapars{n + 1} over Gammapars{x + n + 1}}
      phantom{= mrm{B}pars{x,n + 1},,,,,,,,,,,,}
      pars{~Gamma: Gamma Function~}
      \[5mm] = &
      {n! over Gammapars{x + n + 1}/Gammapars{x}} =
      {n! over x^{overline{n +1}}} =
      bbx{n! over xpars{x + 1}cdotspars{x + n}}
      end{align}






      share|cite|improve this answer























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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        Induction step:



        $$begin{align}
        sum_{k=0}^{n+1}&frac{(-1)^k}{x+k}binom{n+1}k=frac1x+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}left[binom nk+binom n{k-1}right]
        \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}binom{n}{k-1}
        \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-sum_{k=0}^{n-1}frac{(-1)^k}{(x+1)+k}binom{n}{k}
        \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-frac{n!}{(x+1)(x+2)cdots(x+n+1)}+frac{(-1)^n}{x+n+1}
        \&=frac{n!(x+n+1)-n!x}{x(x+1)cdots(x+n+1)}=frac{(n+1)!}{x(x+1)cdots(x+n+1)}
        end{align}$$






        share|cite|improve this answer



























          up vote
          3
          down vote



          accepted










          Induction step:



          $$begin{align}
          sum_{k=0}^{n+1}&frac{(-1)^k}{x+k}binom{n+1}k=frac1x+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}left[binom nk+binom n{k-1}right]
          \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}binom{n}{k-1}
          \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-sum_{k=0}^{n-1}frac{(-1)^k}{(x+1)+k}binom{n}{k}
          \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-frac{n!}{(x+1)(x+2)cdots(x+n+1)}+frac{(-1)^n}{x+n+1}
          \&=frac{n!(x+n+1)-n!x}{x(x+1)cdots(x+n+1)}=frac{(n+1)!}{x(x+1)cdots(x+n+1)}
          end{align}$$






          share|cite|improve this answer

























            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            Induction step:



            $$begin{align}
            sum_{k=0}^{n+1}&frac{(-1)^k}{x+k}binom{n+1}k=frac1x+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}left[binom nk+binom n{k-1}right]
            \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}binom{n}{k-1}
            \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-sum_{k=0}^{n-1}frac{(-1)^k}{(x+1)+k}binom{n}{k}
            \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-frac{n!}{(x+1)(x+2)cdots(x+n+1)}+frac{(-1)^n}{x+n+1}
            \&=frac{n!(x+n+1)-n!x}{x(x+1)cdots(x+n+1)}=frac{(n+1)!}{x(x+1)cdots(x+n+1)}
            end{align}$$






            share|cite|improve this answer














            Induction step:



            $$begin{align}
            sum_{k=0}^{n+1}&frac{(-1)^k}{x+k}binom{n+1}k=frac1x+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}left[binom nk+binom n{k-1}right]
            \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}+sum_{k=1}^{n}frac{(-1)^k}{x+k}binom{n}{k-1}
            \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-sum_{k=0}^{n-1}frac{(-1)^k}{(x+1)+k}binom{n}{k}
            \&=frac{n!}{x(x+1)cdots(x+n)}+frac{(-1)^{n+1}}{x+n+1}-frac{n!}{(x+1)(x+2)cdots(x+n+1)}+frac{(-1)^n}{x+n+1}
            \&=frac{n!(x+n+1)-n!x}{x(x+1)cdots(x+n+1)}=frac{(n+1)!}{x(x+1)cdots(x+n+1)}
            end{align}$$







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            edited Nov 27 at 19:39









            darij grinberg

            10.2k33061




            10.2k33061










            answered Nov 19 at 16:35









            ajotatxe

            53k23890




            53k23890






















                up vote
                4
                down vote













                Consider the (unique) polynomial $p(x)inmathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,ldots,n$. Clearly, $p(x)$ is the constant polynomial $1$.



                However, using Lagrange interpolation, we have
                $$p(x)=sum_{k=0}^n,p(-k),frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)},,$$
                where $[n]:={0,1,2,ldots,n}$. This means
                $$1=sum_{k=0}^n,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)}=sum_{k=0}^n,(-1)^k,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{k!,(n-k)!},.$$
                Multiplying both sides by $dfrac{n!}{prodlimits_{jin[n]},(x+j)}$ yields
                $$frac{n!}{prodlimits_{j=0}^n,(x+j)}=sum_{k=0}^n,(-1)^k,left(frac{n!}{k!,(n-k)!}right),frac{1}{x+k}=sum_{k=0}^n,binom{n}{k},frac{(-1)^k}{x+k},.$$






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                  up vote
                  4
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                  Consider the (unique) polynomial $p(x)inmathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,ldots,n$. Clearly, $p(x)$ is the constant polynomial $1$.



                  However, using Lagrange interpolation, we have
                  $$p(x)=sum_{k=0}^n,p(-k),frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)},,$$
                  where $[n]:={0,1,2,ldots,n}$. This means
                  $$1=sum_{k=0}^n,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)}=sum_{k=0}^n,(-1)^k,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{k!,(n-k)!},.$$
                  Multiplying both sides by $dfrac{n!}{prodlimits_{jin[n]},(x+j)}$ yields
                  $$frac{n!}{prodlimits_{j=0}^n,(x+j)}=sum_{k=0}^n,(-1)^k,left(frac{n!}{k!,(n-k)!}right),frac{1}{x+k}=sum_{k=0}^n,binom{n}{k},frac{(-1)^k}{x+k},.$$






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                    up vote
                    4
                    down vote










                    up vote
                    4
                    down vote









                    Consider the (unique) polynomial $p(x)inmathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,ldots,n$. Clearly, $p(x)$ is the constant polynomial $1$.



                    However, using Lagrange interpolation, we have
                    $$p(x)=sum_{k=0}^n,p(-k),frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)},,$$
                    where $[n]:={0,1,2,ldots,n}$. This means
                    $$1=sum_{k=0}^n,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)}=sum_{k=0}^n,(-1)^k,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{k!,(n-k)!},.$$
                    Multiplying both sides by $dfrac{n!}{prodlimits_{jin[n]},(x+j)}$ yields
                    $$frac{n!}{prodlimits_{j=0}^n,(x+j)}=sum_{k=0}^n,(-1)^k,left(frac{n!}{k!,(n-k)!}right),frac{1}{x+k}=sum_{k=0}^n,binom{n}{k},frac{(-1)^k}{x+k},.$$






                    share|cite|improve this answer














                    Consider the (unique) polynomial $p(x)inmathbb{Q}[x]$ of degree at most $n$ such that $p(-k)=1$ for all $k=0,1,2,ldots,n$. Clearly, $p(x)$ is the constant polynomial $1$.



                    However, using Lagrange interpolation, we have
                    $$p(x)=sum_{k=0}^n,p(-k),frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)},,$$
                    where $[n]:={0,1,2,ldots,n}$. This means
                    $$1=sum_{k=0}^n,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{prodlimits_{jin[n]setminus{k}},(-k+j)}=sum_{k=0}^n,(-1)^k,frac{prodlimits_{jin[n]setminus{k}},(x+j)}{k!,(n-k)!},.$$
                    Multiplying both sides by $dfrac{n!}{prodlimits_{jin[n]},(x+j)}$ yields
                    $$frac{n!}{prodlimits_{j=0}^n,(x+j)}=sum_{k=0}^n,(-1)^k,left(frac{n!}{k!,(n-k)!}right),frac{1}{x+k}=sum_{k=0}^n,binom{n}{k},frac{(-1)^k}{x+k},.$$







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                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 27 at 19:42









                    darij grinberg

                    10.2k33061




                    10.2k33061










                    answered Nov 22 at 17:11









                    Batominovski

                    33.5k33292




                    33.5k33292






















                        up vote
                        3
                        down vote













                        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                        newcommand{dd}{mathrm{d}}
                        newcommand{ds}[1]{displaystyle{#1}}
                        newcommand{expo}[1]{,mathrm{e}^{#1},}
                        newcommand{ic}{mathrm{i}}
                        newcommand{mc}[1]{mathcal{#1}}
                        newcommand{mrm}[1]{mathrm{#1}}
                        newcommand{pars}[1]{left(,{#1},right)}
                        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                        newcommand{verts}[1]{leftvert,{#1},rightvert}$




                        With $ds{Repars{x} > 0}$:




                        begin{align}
                        &bbox[10px,#ffd]{sum_{k = 0}^{n}{pars{-1}^{k} over k + x}{n choose k}} =
                        sum_{k = 0}^{n}pars{-1}^{k}
                        pars{int_{0}^{1}t^{k + x - 1},dd t}{n choose k}
                        \[5mm] = &
                        int_{0}^{1}t^{x - 1}sum_{k = 0}^{n}
                        {n choose k}pars{-t}^{k},dd t
                        \[5mm] = &
                        int_{0}^{1}t^{x - 1},pars{1 - t}^{n},dd t =
                        mrm{B}pars{x,n + 1} pars{~mrm{B}: Beta Function~}
                        \[5mm] = &
                        {Gammapars{x}Gammapars{n + 1} over Gammapars{x + n + 1}}
                        phantom{= mrm{B}pars{x,n + 1},,,,,,,,,,,,}
                        pars{~Gamma: Gamma Function~}
                        \[5mm] = &
                        {n! over Gammapars{x + n + 1}/Gammapars{x}} =
                        {n! over x^{overline{n +1}}} =
                        bbx{n! over xpars{x + 1}cdotspars{x + n}}
                        end{align}






                        share|cite|improve this answer



























                          up vote
                          3
                          down vote













                          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                          newcommand{dd}{mathrm{d}}
                          newcommand{ds}[1]{displaystyle{#1}}
                          newcommand{expo}[1]{,mathrm{e}^{#1},}
                          newcommand{ic}{mathrm{i}}
                          newcommand{mc}[1]{mathcal{#1}}
                          newcommand{mrm}[1]{mathrm{#1}}
                          newcommand{pars}[1]{left(,{#1},right)}
                          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                          newcommand{verts}[1]{leftvert,{#1},rightvert}$




                          With $ds{Repars{x} > 0}$:




                          begin{align}
                          &bbox[10px,#ffd]{sum_{k = 0}^{n}{pars{-1}^{k} over k + x}{n choose k}} =
                          sum_{k = 0}^{n}pars{-1}^{k}
                          pars{int_{0}^{1}t^{k + x - 1},dd t}{n choose k}
                          \[5mm] = &
                          int_{0}^{1}t^{x - 1}sum_{k = 0}^{n}
                          {n choose k}pars{-t}^{k},dd t
                          \[5mm] = &
                          int_{0}^{1}t^{x - 1},pars{1 - t}^{n},dd t =
                          mrm{B}pars{x,n + 1} pars{~mrm{B}: Beta Function~}
                          \[5mm] = &
                          {Gammapars{x}Gammapars{n + 1} over Gammapars{x + n + 1}}
                          phantom{= mrm{B}pars{x,n + 1},,,,,,,,,,,,}
                          pars{~Gamma: Gamma Function~}
                          \[5mm] = &
                          {n! over Gammapars{x + n + 1}/Gammapars{x}} =
                          {n! over x^{overline{n +1}}} =
                          bbx{n! over xpars{x + 1}cdotspars{x + n}}
                          end{align}






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                            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                            newcommand{dd}{mathrm{d}}
                            newcommand{ds}[1]{displaystyle{#1}}
                            newcommand{expo}[1]{,mathrm{e}^{#1},}
                            newcommand{ic}{mathrm{i}}
                            newcommand{mc}[1]{mathcal{#1}}
                            newcommand{mrm}[1]{mathrm{#1}}
                            newcommand{pars}[1]{left(,{#1},right)}
                            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                            newcommand{verts}[1]{leftvert,{#1},rightvert}$




                            With $ds{Repars{x} > 0}$:




                            begin{align}
                            &bbox[10px,#ffd]{sum_{k = 0}^{n}{pars{-1}^{k} over k + x}{n choose k}} =
                            sum_{k = 0}^{n}pars{-1}^{k}
                            pars{int_{0}^{1}t^{k + x - 1},dd t}{n choose k}
                            \[5mm] = &
                            int_{0}^{1}t^{x - 1}sum_{k = 0}^{n}
                            {n choose k}pars{-t}^{k},dd t
                            \[5mm] = &
                            int_{0}^{1}t^{x - 1},pars{1 - t}^{n},dd t =
                            mrm{B}pars{x,n + 1} pars{~mrm{B}: Beta Function~}
                            \[5mm] = &
                            {Gammapars{x}Gammapars{n + 1} over Gammapars{x + n + 1}}
                            phantom{= mrm{B}pars{x,n + 1},,,,,,,,,,,,}
                            pars{~Gamma: Gamma Function~}
                            \[5mm] = &
                            {n! over Gammapars{x + n + 1}/Gammapars{x}} =
                            {n! over x^{overline{n +1}}} =
                            bbx{n! over xpars{x + 1}cdotspars{x + n}}
                            end{align}






                            share|cite|improve this answer














                            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                            newcommand{dd}{mathrm{d}}
                            newcommand{ds}[1]{displaystyle{#1}}
                            newcommand{expo}[1]{,mathrm{e}^{#1},}
                            newcommand{ic}{mathrm{i}}
                            newcommand{mc}[1]{mathcal{#1}}
                            newcommand{mrm}[1]{mathrm{#1}}
                            newcommand{pars}[1]{left(,{#1},right)}
                            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                            newcommand{verts}[1]{leftvert,{#1},rightvert}$




                            With $ds{Repars{x} > 0}$:




                            begin{align}
                            &bbox[10px,#ffd]{sum_{k = 0}^{n}{pars{-1}^{k} over k + x}{n choose k}} =
                            sum_{k = 0}^{n}pars{-1}^{k}
                            pars{int_{0}^{1}t^{k + x - 1},dd t}{n choose k}
                            \[5mm] = &
                            int_{0}^{1}t^{x - 1}sum_{k = 0}^{n}
                            {n choose k}pars{-t}^{k},dd t
                            \[5mm] = &
                            int_{0}^{1}t^{x - 1},pars{1 - t}^{n},dd t =
                            mrm{B}pars{x,n + 1} pars{~mrm{B}: Beta Function~}
                            \[5mm] = &
                            {Gammapars{x}Gammapars{n + 1} over Gammapars{x + n + 1}}
                            phantom{= mrm{B}pars{x,n + 1},,,,,,,,,,,,}
                            pars{~Gamma: Gamma Function~}
                            \[5mm] = &
                            {n! over Gammapars{x + n + 1}/Gammapars{x}} =
                            {n! over x^{overline{n +1}}} =
                            bbx{n! over xpars{x + 1}cdotspars{x + n}}
                            end{align}







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 28 at 0:44

























                            answered Nov 27 at 17:43









                            Felix Marin

                            66.8k7107139




                            66.8k7107139






























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