Magic relation for harmonic numbers












3












$begingroup$


I was doing same computations for an exercise and I came up with the following relation



$$sum_{k=1}^nfrac{1}{k}=sum_{j=1}^nfrac{1}{(j-1)!}sum_{i_1+dots+i_j=n\i_1,dots i_jgeq1}frac{1}{i_1i_2dots i_j}$$



for all $ngeq1.$



Any idea how to prove it algebraically?



Small result: I'm sure that the two sums are not equal terms by terms.






sequences-and-series combinatorics power-series







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    I was doing same computations for an exercise and I came up with the following relation



    $$sum_{k=1}^nfrac{1}{k}=sum_{j=1}^nfrac{1}{(j-1)!}sum_{i_1+dots+i_j=n\i_1,dots i_jgeq1}frac{1}{i_1i_2dots i_j}$$



    for all $ngeq1.$



    Any idea how to prove it algebraically?



    Small result: I'm sure that the two sums are not equal terms by terms.






    sequences-and-series combinatorics power-series







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I was doing same computations for an exercise and I came up with the following relation



      $$sum_{k=1}^nfrac{1}{k}=sum_{j=1}^nfrac{1}{(j-1)!}sum_{i_1+dots+i_j=n\i_1,dots i_jgeq1}frac{1}{i_1i_2dots i_j}$$



      for all $ngeq1.$



      Any idea how to prove it algebraically?



      Small result: I'm sure that the two sums are not equal terms by terms.






      sequences-and-series combinatorics power-series







      share|cite|improve this question









      $endgroup$




      I was doing same computations for an exercise and I came up with the following relation



      $$sum_{k=1}^nfrac{1}{k}=sum_{j=1}^nfrac{1}{(j-1)!}sum_{i_1+dots+i_j=n\i_1,dots i_jgeq1}frac{1}{i_1i_2dots i_j}$$



      for all $ngeq1.$



      Any idea how to prove it algebraically?



      Small result: I'm sure that the two sums are not equal terms by terms.






      sequences-and-series combinatorics power-series




      sequences-and-series combinatorics power-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 27 '18 at 23:47









      bojicabojica

      251114




      251114






















          1 Answer
          1






          active

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          2












          $begingroup$

          We have using formal power series that the RHS is



          $$sum_{j=1}^n frac{1}{(j-1)!}
          [z^n] left(logfrac{1}{1-z}right)^j
          = [z^n] sum_{j=1}^n frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j.$$



          Now since $logfrac{1}{1-z} = z + cdots$ we may extend $j$ beyond $n$
          without contributing to the coefficient extractor:



          $$[z^n] sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j
          \ = [z^n] logfrac{1}{1-z} sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^{j-1}
          \ = [z^n] logfrac{1}{1-z}
          exp logfrac{1}{1-z}
          = [z^n] frac{1}{1-z} logfrac{1}{1-z}
          = sum_{k=1}^n frac{1}{k} = H_n.$$



          Remark. Restricting the proof to combinatorial methods we have for
          the combinatorial class $mathcal{P}$ of permutations the
          specification as sets of labeled cycles, which is



          $$deftextsc#1{dosc#1csod}
          defdosc#1#2csod{{rm #1{small #2}}}mathcal{P} =
          textsc{SET}(textsc{CYC}(mathcal{Z}))$$



          Since permutations have EGF $sum_{nge 0} n! z^n/n! = frac{1}{1-z}$
          we have the identity



          $$frac{1}{1-z} = exp logfrac{1}{1-z}$$



          that was used above.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Elegant solution! (+1)
            $endgroup$
            – Markus Scheuer
            Nov 28 '18 at 20:27











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

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          active

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          2












          $begingroup$

          We have using formal power series that the RHS is



          $$sum_{j=1}^n frac{1}{(j-1)!}
          [z^n] left(logfrac{1}{1-z}right)^j
          = [z^n] sum_{j=1}^n frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j.$$



          Now since $logfrac{1}{1-z} = z + cdots$ we may extend $j$ beyond $n$
          without contributing to the coefficient extractor:



          $$[z^n] sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j
          \ = [z^n] logfrac{1}{1-z} sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^{j-1}
          \ = [z^n] logfrac{1}{1-z}
          exp logfrac{1}{1-z}
          = [z^n] frac{1}{1-z} logfrac{1}{1-z}
          = sum_{k=1}^n frac{1}{k} = H_n.$$



          Remark. Restricting the proof to combinatorial methods we have for
          the combinatorial class $mathcal{P}$ of permutations the
          specification as sets of labeled cycles, which is



          $$deftextsc#1{dosc#1csod}
          defdosc#1#2csod{{rm #1{small #2}}}mathcal{P} =
          textsc{SET}(textsc{CYC}(mathcal{Z}))$$



          Since permutations have EGF $sum_{nge 0} n! z^n/n! = frac{1}{1-z}$
          we have the identity



          $$frac{1}{1-z} = exp logfrac{1}{1-z}$$



          that was used above.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Elegant solution! (+1)
            $endgroup$
            – Markus Scheuer
            Nov 28 '18 at 20:27
















          2












          $begingroup$

          We have using formal power series that the RHS is



          $$sum_{j=1}^n frac{1}{(j-1)!}
          [z^n] left(logfrac{1}{1-z}right)^j
          = [z^n] sum_{j=1}^n frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j.$$



          Now since $logfrac{1}{1-z} = z + cdots$ we may extend $j$ beyond $n$
          without contributing to the coefficient extractor:



          $$[z^n] sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j
          \ = [z^n] logfrac{1}{1-z} sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^{j-1}
          \ = [z^n] logfrac{1}{1-z}
          exp logfrac{1}{1-z}
          = [z^n] frac{1}{1-z} logfrac{1}{1-z}
          = sum_{k=1}^n frac{1}{k} = H_n.$$



          Remark. Restricting the proof to combinatorial methods we have for
          the combinatorial class $mathcal{P}$ of permutations the
          specification as sets of labeled cycles, which is



          $$deftextsc#1{dosc#1csod}
          defdosc#1#2csod{{rm #1{small #2}}}mathcal{P} =
          textsc{SET}(textsc{CYC}(mathcal{Z}))$$



          Since permutations have EGF $sum_{nge 0} n! z^n/n! = frac{1}{1-z}$
          we have the identity



          $$frac{1}{1-z} = exp logfrac{1}{1-z}$$



          that was used above.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Elegant solution! (+1)
            $endgroup$
            – Markus Scheuer
            Nov 28 '18 at 20:27














          2












          2








          2





          $begingroup$

          We have using formal power series that the RHS is



          $$sum_{j=1}^n frac{1}{(j-1)!}
          [z^n] left(logfrac{1}{1-z}right)^j
          = [z^n] sum_{j=1}^n frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j.$$



          Now since $logfrac{1}{1-z} = z + cdots$ we may extend $j$ beyond $n$
          without contributing to the coefficient extractor:



          $$[z^n] sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j
          \ = [z^n] logfrac{1}{1-z} sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^{j-1}
          \ = [z^n] logfrac{1}{1-z}
          exp logfrac{1}{1-z}
          = [z^n] frac{1}{1-z} logfrac{1}{1-z}
          = sum_{k=1}^n frac{1}{k} = H_n.$$



          Remark. Restricting the proof to combinatorial methods we have for
          the combinatorial class $mathcal{P}$ of permutations the
          specification as sets of labeled cycles, which is



          $$deftextsc#1{dosc#1csod}
          defdosc#1#2csod{{rm #1{small #2}}}mathcal{P} =
          textsc{SET}(textsc{CYC}(mathcal{Z}))$$



          Since permutations have EGF $sum_{nge 0} n! z^n/n! = frac{1}{1-z}$
          we have the identity



          $$frac{1}{1-z} = exp logfrac{1}{1-z}$$



          that was used above.






          share|cite|improve this answer











          $endgroup$



          We have using formal power series that the RHS is



          $$sum_{j=1}^n frac{1}{(j-1)!}
          [z^n] left(logfrac{1}{1-z}right)^j
          = [z^n] sum_{j=1}^n frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j.$$



          Now since $logfrac{1}{1-z} = z + cdots$ we may extend $j$ beyond $n$
          without contributing to the coefficient extractor:



          $$[z^n] sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^j
          \ = [z^n] logfrac{1}{1-z} sum_{jge 1} frac{1}{(j-1)!}
          left(logfrac{1}{1-z}right)^{j-1}
          \ = [z^n] logfrac{1}{1-z}
          exp logfrac{1}{1-z}
          = [z^n] frac{1}{1-z} logfrac{1}{1-z}
          = sum_{k=1}^n frac{1}{k} = H_n.$$



          Remark. Restricting the proof to combinatorial methods we have for
          the combinatorial class $mathcal{P}$ of permutations the
          specification as sets of labeled cycles, which is



          $$deftextsc#1{dosc#1csod}
          defdosc#1#2csod{{rm #1{small #2}}}mathcal{P} =
          textsc{SET}(textsc{CYC}(mathcal{Z}))$$



          Since permutations have EGF $sum_{nge 0} n! z^n/n! = frac{1}{1-z}$
          we have the identity



          $$frac{1}{1-z} = exp logfrac{1}{1-z}$$



          that was used above.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 '18 at 15:39

























          answered Nov 28 '18 at 18:13









          Marko RiedelMarko Riedel

          39.8k339108




          39.8k339108








          • 1




            $begingroup$
            Elegant solution! (+1)
            $endgroup$
            – Markus Scheuer
            Nov 28 '18 at 20:27














          • 1




            $begingroup$
            Elegant solution! (+1)
            $endgroup$
            – Markus Scheuer
            Nov 28 '18 at 20:27








          1




          1




          $begingroup$
          Elegant solution! (+1)
          $endgroup$
          – Markus Scheuer
          Nov 28 '18 at 20:27




          $begingroup$
          Elegant solution! (+1)
          $endgroup$
          – Markus Scheuer
          Nov 28 '18 at 20:27


















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