Showing $sumlimits_{j=0}^M frac{M choose j}{N+M choose j} = frac{N+M+1}{N+1}$












5












$begingroup$


In an answer to another question, I stated $$sumlimits_{j=0}^M frac{M choose j}{N+M choose j} = frac{N+M+1}{N+1}.$$



It is clearly true when $N=0$ since you add up $M+1$ copies of $1$, and when $M=0$ since you add up one copy of $1$. And, for example, with $M=4$ and $N=9$ you get $frac{1}{1}+frac{4}{13}+frac{6}{78}+frac{4}{286}+frac{1}{715} = frac{14}{10}$ as expected.



But how might you approach a general proof?






combinatorics binomial-coefficients







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    In an answer to another question, I stated $$sumlimits_{j=0}^M frac{M choose j}{N+M choose j} = frac{N+M+1}{N+1}.$$



    It is clearly true when $N=0$ since you add up $M+1$ copies of $1$, and when $M=0$ since you add up one copy of $1$. And, for example, with $M=4$ and $N=9$ you get $frac{1}{1}+frac{4}{13}+frac{6}{78}+frac{4}{286}+frac{1}{715} = frac{14}{10}$ as expected.



    But how might you approach a general proof?






    combinatorics binomial-coefficients







    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      In an answer to another question, I stated $$sumlimits_{j=0}^M frac{M choose j}{N+M choose j} = frac{N+M+1}{N+1}.$$



      It is clearly true when $N=0$ since you add up $M+1$ copies of $1$, and when $M=0$ since you add up one copy of $1$. And, for example, with $M=4$ and $N=9$ you get $frac{1}{1}+frac{4}{13}+frac{6}{78}+frac{4}{286}+frac{1}{715} = frac{14}{10}$ as expected.



      But how might you approach a general proof?






      combinatorics binomial-coefficients







      share|cite|improve this question











      $endgroup$




      In an answer to another question, I stated $$sumlimits_{j=0}^M frac{M choose j}{N+M choose j} = frac{N+M+1}{N+1}.$$



      It is clearly true when $N=0$ since you add up $M+1$ copies of $1$, and when $M=0$ since you add up one copy of $1$. And, for example, with $M=4$ and $N=9$ you get $frac{1}{1}+frac{4}{13}+frac{6}{78}+frac{4}{286}+frac{1}{715} = frac{14}{10}$ as expected.



      But how might you approach a general proof?






      combinatorics binomial-coefficients




      combinatorics binomial-coefficients






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 '18 at 13:23









      Robert Z

      94.5k1063134




      94.5k1063134










      asked Nov 25 '18 at 13:11









      HenryHenry

      99.1k478164




      99.1k478164






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Hint. Note that
          $$frac{M choose j}{N+M choose j}=frac{binom{N+M-j}{N}}{binom{N+M}{N}}.$$
          Hence, we can rewrite the sum as
          $$sum_{j=0}^M frac{M choose j}{N+M choose j}=frac{1}{binom{N+M}{N}}sum_{j=0}^M binom{N+M-j}{N}=frac{1}{binom{N+M}{N}}sum_{i=N}^{N+M} binom{i}{N}.$$
          Finally use the Hockey-stick identity.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That seems to work well. In my example of $M=4,N=9$ this represents $frac{715}{715}+frac{220}{715}+frac{55}{715}+frac{10}{715}+frac{1}{715} = frac{1001}{715}=frac{14 choose 10}{13 choose 9}=frac{14}{10}$
            $endgroup$
            – Henry
            Nov 25 '18 at 19:09










          • $begingroup$
            @Henry Well done!!
            $endgroup$
            – Robert Z
            Nov 25 '18 at 19:19



















          1












          $begingroup$

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




          $ds{sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} =
          {N + M + 1 over N + 1}: {LARGE ?}}$          .




          begin{align}
          sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} & =
          sum_{j = 0}^{M}{M!/bracks{j!pars{M - j}!} over
          pars{N + M}!/bracks{j!pars{N + M - j}!}}
          \[5mm] & =
          {M!, N! over pars{N + M}!}sum_{j = 0}^{M}
          {N + M - j choose M - j}
          \[5mm] & =
          {M!, N! over pars{N + M}!}pars{-1}^{M}
          sum_{j = 0}^{M}pars{-1}^{j}
          bracks{z^{M - j}}pars{1 + z}^{-N - 1}
          \[5mm] & =
          {M!, N! over pars{N + M}!}pars{-1}^{M}
          bracks{z^{M}}pars{1 + z}^{-N - 1},
          {pars{-z}^{M + 1} - 1 over pars{-z} - 1}
          \[5mm] & =
          {M!, N! over pars{N + M}!}pars{-1}^{M}
          bracks{z^{M}}pars{1 + z}^{-N - 2}
          \[5mm] & =
          {M!, N! over pars{N + M}!}pars{-1}^{M}
          braces{{-bracks{-N - 2} + M - 1 choose M}pars{-1}^{M}}
          \[5mm] & = bbx{N + M + 1 over N + 1}
          end{align}






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            2 Answers
            2






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            oldest

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            active

            oldest

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            6












            $begingroup$

            Hint. Note that
            $$frac{M choose j}{N+M choose j}=frac{binom{N+M-j}{N}}{binom{N+M}{N}}.$$
            Hence, we can rewrite the sum as
            $$sum_{j=0}^M frac{M choose j}{N+M choose j}=frac{1}{binom{N+M}{N}}sum_{j=0}^M binom{N+M-j}{N}=frac{1}{binom{N+M}{N}}sum_{i=N}^{N+M} binom{i}{N}.$$
            Finally use the Hockey-stick identity.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That seems to work well. In my example of $M=4,N=9$ this represents $frac{715}{715}+frac{220}{715}+frac{55}{715}+frac{10}{715}+frac{1}{715} = frac{1001}{715}=frac{14 choose 10}{13 choose 9}=frac{14}{10}$
              $endgroup$
              – Henry
              Nov 25 '18 at 19:09










            • $begingroup$
              @Henry Well done!!
              $endgroup$
              – Robert Z
              Nov 25 '18 at 19:19
















            6












            $begingroup$

            Hint. Note that
            $$frac{M choose j}{N+M choose j}=frac{binom{N+M-j}{N}}{binom{N+M}{N}}.$$
            Hence, we can rewrite the sum as
            $$sum_{j=0}^M frac{M choose j}{N+M choose j}=frac{1}{binom{N+M}{N}}sum_{j=0}^M binom{N+M-j}{N}=frac{1}{binom{N+M}{N}}sum_{i=N}^{N+M} binom{i}{N}.$$
            Finally use the Hockey-stick identity.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              That seems to work well. In my example of $M=4,N=9$ this represents $frac{715}{715}+frac{220}{715}+frac{55}{715}+frac{10}{715}+frac{1}{715} = frac{1001}{715}=frac{14 choose 10}{13 choose 9}=frac{14}{10}$
              $endgroup$
              – Henry
              Nov 25 '18 at 19:09










            • $begingroup$
              @Henry Well done!!
              $endgroup$
              – Robert Z
              Nov 25 '18 at 19:19














            6












            6








            6





            $begingroup$

            Hint. Note that
            $$frac{M choose j}{N+M choose j}=frac{binom{N+M-j}{N}}{binom{N+M}{N}}.$$
            Hence, we can rewrite the sum as
            $$sum_{j=0}^M frac{M choose j}{N+M choose j}=frac{1}{binom{N+M}{N}}sum_{j=0}^M binom{N+M-j}{N}=frac{1}{binom{N+M}{N}}sum_{i=N}^{N+M} binom{i}{N}.$$
            Finally use the Hockey-stick identity.






            share|cite|improve this answer











            $endgroup$



            Hint. Note that
            $$frac{M choose j}{N+M choose j}=frac{binom{N+M-j}{N}}{binom{N+M}{N}}.$$
            Hence, we can rewrite the sum as
            $$sum_{j=0}^M frac{M choose j}{N+M choose j}=frac{1}{binom{N+M}{N}}sum_{j=0}^M binom{N+M-j}{N}=frac{1}{binom{N+M}{N}}sum_{i=N}^{N+M} binom{i}{N}.$$
            Finally use the Hockey-stick identity.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 25 '18 at 19:19

























            answered Nov 25 '18 at 13:16









            Robert ZRobert Z

            94.5k1063134




            94.5k1063134












            • $begingroup$
              That seems to work well. In my example of $M=4,N=9$ this represents $frac{715}{715}+frac{220}{715}+frac{55}{715}+frac{10}{715}+frac{1}{715} = frac{1001}{715}=frac{14 choose 10}{13 choose 9}=frac{14}{10}$
              $endgroup$
              – Henry
              Nov 25 '18 at 19:09










            • $begingroup$
              @Henry Well done!!
              $endgroup$
              – Robert Z
              Nov 25 '18 at 19:19


















            • $begingroup$
              That seems to work well. In my example of $M=4,N=9$ this represents $frac{715}{715}+frac{220}{715}+frac{55}{715}+frac{10}{715}+frac{1}{715} = frac{1001}{715}=frac{14 choose 10}{13 choose 9}=frac{14}{10}$
              $endgroup$
              – Henry
              Nov 25 '18 at 19:09










            • $begingroup$
              @Henry Well done!!
              $endgroup$
              – Robert Z
              Nov 25 '18 at 19:19
















            $begingroup$
            That seems to work well. In my example of $M=4,N=9$ this represents $frac{715}{715}+frac{220}{715}+frac{55}{715}+frac{10}{715}+frac{1}{715} = frac{1001}{715}=frac{14 choose 10}{13 choose 9}=frac{14}{10}$
            $endgroup$
            – Henry
            Nov 25 '18 at 19:09




            $begingroup$
            That seems to work well. In my example of $M=4,N=9$ this represents $frac{715}{715}+frac{220}{715}+frac{55}{715}+frac{10}{715}+frac{1}{715} = frac{1001}{715}=frac{14 choose 10}{13 choose 9}=frac{14}{10}$
            $endgroup$
            – Henry
            Nov 25 '18 at 19:09












            $begingroup$
            @Henry Well done!!
            $endgroup$
            – Robert Z
            Nov 25 '18 at 19:19




            $begingroup$
            @Henry Well done!!
            $endgroup$
            – Robert Z
            Nov 25 '18 at 19:19











            1












            $begingroup$

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




            $ds{sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} =
            {N + M + 1 over N + 1}: {LARGE ?}}$            .




            begin{align}
            sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} & =
            sum_{j = 0}^{M}{M!/bracks{j!pars{M - j}!} over
            pars{N + M}!/bracks{j!pars{N + M - j}!}}
            \[5mm] & =
            {M!, N! over pars{N + M}!}sum_{j = 0}^{M}
            {N + M - j choose M - j}
            \[5mm] & =
            {M!, N! over pars{N + M}!}pars{-1}^{M}
            sum_{j = 0}^{M}pars{-1}^{j}
            bracks{z^{M - j}}pars{1 + z}^{-N - 1}
            \[5mm] & =
            {M!, N! over pars{N + M}!}pars{-1}^{M}
            bracks{z^{M}}pars{1 + z}^{-N - 1},
            {pars{-z}^{M + 1} - 1 over pars{-z} - 1}
            \[5mm] & =
            {M!, N! over pars{N + M}!}pars{-1}^{M}
            bracks{z^{M}}pars{1 + z}^{-N - 2}
            \[5mm] & =
            {M!, N! over pars{N + M}!}pars{-1}^{M}
            braces{{-bracks{-N - 2} + M - 1 choose M}pars{-1}^{M}}
            \[5mm] & = bbx{N + M + 1 over N + 1}
            end{align}






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

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




              $ds{sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} =
              {N + M + 1 over N + 1}: {LARGE ?}}$              .




              begin{align}
              sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} & =
              sum_{j = 0}^{M}{M!/bracks{j!pars{M - j}!} over
              pars{N + M}!/bracks{j!pars{N + M - j}!}}
              \[5mm] & =
              {M!, N! over pars{N + M}!}sum_{j = 0}^{M}
              {N + M - j choose M - j}
              \[5mm] & =
              {M!, N! over pars{N + M}!}pars{-1}^{M}
              sum_{j = 0}^{M}pars{-1}^{j}
              bracks{z^{M - j}}pars{1 + z}^{-N - 1}
              \[5mm] & =
              {M!, N! over pars{N + M}!}pars{-1}^{M}
              bracks{z^{M}}pars{1 + z}^{-N - 1},
              {pars{-z}^{M + 1} - 1 over pars{-z} - 1}
              \[5mm] & =
              {M!, N! over pars{N + M}!}pars{-1}^{M}
              bracks{z^{M}}pars{1 + z}^{-N - 2}
              \[5mm] & =
              {M!, N! over pars{N + M}!}pars{-1}^{M}
              braces{{-bracks{-N - 2} + M - 1 choose M}pars{-1}^{M}}
              \[5mm] & = bbx{N + M + 1 over N + 1}
              end{align}






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

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




                $ds{sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} =
                {N + M + 1 over N + 1}: {LARGE ?}}$                .




                begin{align}
                sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} & =
                sum_{j = 0}^{M}{M!/bracks{j!pars{M - j}!} over
                pars{N + M}!/bracks{j!pars{N + M - j}!}}
                \[5mm] & =
                {M!, N! over pars{N + M}!}sum_{j = 0}^{M}
                {N + M - j choose M - j}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                sum_{j = 0}^{M}pars{-1}^{j}
                bracks{z^{M - j}}pars{1 + z}^{-N - 1}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                bracks{z^{M}}pars{1 + z}^{-N - 1},
                {pars{-z}^{M + 1} - 1 over pars{-z} - 1}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                bracks{z^{M}}pars{1 + z}^{-N - 2}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                braces{{-bracks{-N - 2} + M - 1 choose M}pars{-1}^{M}}
                \[5mm] & = bbx{N + M + 1 over N + 1}
                end{align}






                share|cite|improve this answer











                $endgroup$



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




                $ds{sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} =
                {N + M + 1 over N + 1}: {LARGE ?}}$                .




                begin{align}
                sum_{j = 0}^{M}{{M choose j} over {N + M choose j}} & =
                sum_{j = 0}^{M}{M!/bracks{j!pars{M - j}!} over
                pars{N + M}!/bracks{j!pars{N + M - j}!}}
                \[5mm] & =
                {M!, N! over pars{N + M}!}sum_{j = 0}^{M}
                {N + M - j choose M - j}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                sum_{j = 0}^{M}pars{-1}^{j}
                bracks{z^{M - j}}pars{1 + z}^{-N - 1}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                bracks{z^{M}}pars{1 + z}^{-N - 1},
                {pars{-z}^{M + 1} - 1 over pars{-z} - 1}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                bracks{z^{M}}pars{1 + z}^{-N - 2}
                \[5mm] & =
                {M!, N! over pars{N + M}!}pars{-1}^{M}
                braces{{-bracks{-N - 2} + M - 1 choose M}pars{-1}^{M}}
                \[5mm] & = bbx{N + M + 1 over N + 1}
                end{align}







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 28 '18 at 20:50

























                answered Nov 26 '18 at 21:55









                Felix MarinFelix Marin

                67.4k7107141




                67.4k7107141






























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                    How to change which sound is reproduced for terminal bell?

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                    3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
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                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                    up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

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                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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