Partial sum of harmonic series and one one function












1












$begingroup$


Suppose that $A$ is a set of cardinality, say $n$. Let $$H_n = 1+1/2+dots +1/n.$$
If $phi:Ato {1,2,dots,n}$ is one-one, then obviously $$sum_{ain A}phi(a)^{-1} = H_n$$
I'm wondering if the converse of the above statement is true, that is, if $phi:Ato {1,2,dots,n}$ is such that $$sum_{ain A}phi(a)^{-1} = H_n,$$ then should $phi$ be necessarily one-one?



It looks to be true at least for small integers say $n=2,3,4$. I have feel that this maybe true. However, I couldn't show this in general. Any hint towards its solution is greatly appreciated.






real-analysis sequences-and-series real-numbers







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    1












    $begingroup$


    Suppose that $A$ is a set of cardinality, say $n$. Let $$H_n = 1+1/2+dots +1/n.$$
    If $phi:Ato {1,2,dots,n}$ is one-one, then obviously $$sum_{ain A}phi(a)^{-1} = H_n$$
    I'm wondering if the converse of the above statement is true, that is, if $phi:Ato {1,2,dots,n}$ is such that $$sum_{ain A}phi(a)^{-1} = H_n,$$ then should $phi$ be necessarily one-one?



    It looks to be true at least for small integers say $n=2,3,4$. I have feel that this maybe true. However, I couldn't show this in general. Any hint towards its solution is greatly appreciated.






    real-analysis sequences-and-series real-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose that $A$ is a set of cardinality, say $n$. Let $$H_n = 1+1/2+dots +1/n.$$
      If $phi:Ato {1,2,dots,n}$ is one-one, then obviously $$sum_{ain A}phi(a)^{-1} = H_n$$
      I'm wondering if the converse of the above statement is true, that is, if $phi:Ato {1,2,dots,n}$ is such that $$sum_{ain A}phi(a)^{-1} = H_n,$$ then should $phi$ be necessarily one-one?



      It looks to be true at least for small integers say $n=2,3,4$. I have feel that this maybe true. However, I couldn't show this in general. Any hint towards its solution is greatly appreciated.






      real-analysis sequences-and-series real-numbers







      share|cite|improve this question











      $endgroup$




      Suppose that $A$ is a set of cardinality, say $n$. Let $$H_n = 1+1/2+dots +1/n.$$
      If $phi:Ato {1,2,dots,n}$ is one-one, then obviously $$sum_{ain A}phi(a)^{-1} = H_n$$
      I'm wondering if the converse of the above statement is true, that is, if $phi:Ato {1,2,dots,n}$ is such that $$sum_{ain A}phi(a)^{-1} = H_n,$$ then should $phi$ be necessarily one-one?



      It looks to be true at least for small integers say $n=2,3,4$. I have feel that this maybe true. However, I couldn't show this in general. Any hint towards its solution is greatly appreciated.






      real-analysis sequences-and-series real-numbers




      real-analysis sequences-and-series real-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 7:06







      Kumara

















      asked Dec 7 '18 at 6:32









      KumaraKumara

      216118




      216118






















          1 Answer
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          $begingroup$

          If some $frac{1}{k}$ terms “disappear” in the sum, the same amount of them should be “overlapping” the terms that are present in the sum. We should show that there’s no chance for it to occur. Unfortunately, that is possible:



          $$1+frac{1}{2}+...+frac{1}{20}=1+ frac{1}{2}+ frac{1}{ 3}+ frac{1}{5}+ frac{1}{5}+ frac{1}{6}+ frac{1}{6}+...+ frac{1}{9}+ frac{1}{11}+...+ frac{1}{14}+ frac{1}{16}+...+ frac{1}{20}+ frac{1}{20}$$



          by the equality that



          $$frac{1}{4}+ frac{1}{10}+ frac{1}{15}= frac{1}{5}+ frac{1}{6}+ frac{1}{20}$$



          See the Wikipedia article Egyptian Fraction for more results about the sum of unit fractions.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. Can you give some intuition on how you guessed it?
            $endgroup$
            – Kumara
            Dec 7 '18 at 7:45










          • $begingroup$
            You can read the Wikipedia article I just added to my answer. I took an example directly from there.
            $endgroup$
            – William Sun
            Dec 7 '18 at 7:49











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          $begingroup$

          If some $frac{1}{k}$ terms “disappear” in the sum, the same amount of them should be “overlapping” the terms that are present in the sum. We should show that there’s no chance for it to occur. Unfortunately, that is possible:



          $$1+frac{1}{2}+...+frac{1}{20}=1+ frac{1}{2}+ frac{1}{ 3}+ frac{1}{5}+ frac{1}{5}+ frac{1}{6}+ frac{1}{6}+...+ frac{1}{9}+ frac{1}{11}+...+ frac{1}{14}+ frac{1}{16}+...+ frac{1}{20}+ frac{1}{20}$$



          by the equality that



          $$frac{1}{4}+ frac{1}{10}+ frac{1}{15}= frac{1}{5}+ frac{1}{6}+ frac{1}{20}$$



          See the Wikipedia article Egyptian Fraction for more results about the sum of unit fractions.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. Can you give some intuition on how you guessed it?
            $endgroup$
            – Kumara
            Dec 7 '18 at 7:45










          • $begingroup$
            You can read the Wikipedia article I just added to my answer. I took an example directly from there.
            $endgroup$
            – William Sun
            Dec 7 '18 at 7:49
















          1












          $begingroup$

          If some $frac{1}{k}$ terms “disappear” in the sum, the same amount of them should be “overlapping” the terms that are present in the sum. We should show that there’s no chance for it to occur. Unfortunately, that is possible:



          $$1+frac{1}{2}+...+frac{1}{20}=1+ frac{1}{2}+ frac{1}{ 3}+ frac{1}{5}+ frac{1}{5}+ frac{1}{6}+ frac{1}{6}+...+ frac{1}{9}+ frac{1}{11}+...+ frac{1}{14}+ frac{1}{16}+...+ frac{1}{20}+ frac{1}{20}$$



          by the equality that



          $$frac{1}{4}+ frac{1}{10}+ frac{1}{15}= frac{1}{5}+ frac{1}{6}+ frac{1}{20}$$



          See the Wikipedia article Egyptian Fraction for more results about the sum of unit fractions.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you. Can you give some intuition on how you guessed it?
            $endgroup$
            – Kumara
            Dec 7 '18 at 7:45










          • $begingroup$
            You can read the Wikipedia article I just added to my answer. I took an example directly from there.
            $endgroup$
            – William Sun
            Dec 7 '18 at 7:49














          1












          1








          1





          $begingroup$

          If some $frac{1}{k}$ terms “disappear” in the sum, the same amount of them should be “overlapping” the terms that are present in the sum. We should show that there’s no chance for it to occur. Unfortunately, that is possible:



          $$1+frac{1}{2}+...+frac{1}{20}=1+ frac{1}{2}+ frac{1}{ 3}+ frac{1}{5}+ frac{1}{5}+ frac{1}{6}+ frac{1}{6}+...+ frac{1}{9}+ frac{1}{11}+...+ frac{1}{14}+ frac{1}{16}+...+ frac{1}{20}+ frac{1}{20}$$



          by the equality that



          $$frac{1}{4}+ frac{1}{10}+ frac{1}{15}= frac{1}{5}+ frac{1}{6}+ frac{1}{20}$$



          See the Wikipedia article Egyptian Fraction for more results about the sum of unit fractions.






          share|cite|improve this answer











          $endgroup$



          If some $frac{1}{k}$ terms “disappear” in the sum, the same amount of them should be “overlapping” the terms that are present in the sum. We should show that there’s no chance for it to occur. Unfortunately, that is possible:



          $$1+frac{1}{2}+...+frac{1}{20}=1+ frac{1}{2}+ frac{1}{ 3}+ frac{1}{5}+ frac{1}{5}+ frac{1}{6}+ frac{1}{6}+...+ frac{1}{9}+ frac{1}{11}+...+ frac{1}{14}+ frac{1}{16}+...+ frac{1}{20}+ frac{1}{20}$$



          by the equality that



          $$frac{1}{4}+ frac{1}{10}+ frac{1}{15}= frac{1}{5}+ frac{1}{6}+ frac{1}{20}$$



          See the Wikipedia article Egyptian Fraction for more results about the sum of unit fractions.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 7:47

























          answered Dec 7 '18 at 7:35









          William SunWilliam Sun

          471111




          471111












          • $begingroup$
            Thank you. Can you give some intuition on how you guessed it?
            $endgroup$
            – Kumara
            Dec 7 '18 at 7:45










          • $begingroup$
            You can read the Wikipedia article I just added to my answer. I took an example directly from there.
            $endgroup$
            – William Sun
            Dec 7 '18 at 7:49


















          • $begingroup$
            Thank you. Can you give some intuition on how you guessed it?
            $endgroup$
            – Kumara
            Dec 7 '18 at 7:45










          • $begingroup$
            You can read the Wikipedia article I just added to my answer. I took an example directly from there.
            $endgroup$
            – William Sun
            Dec 7 '18 at 7:49
















          $begingroup$
          Thank you. Can you give some intuition on how you guessed it?
          $endgroup$
          – Kumara
          Dec 7 '18 at 7:45




          $begingroup$
          Thank you. Can you give some intuition on how you guessed it?
          $endgroup$
          – Kumara
          Dec 7 '18 at 7:45












          $begingroup$
          You can read the Wikipedia article I just added to my answer. I took an example directly from there.
          $endgroup$
          – William Sun
          Dec 7 '18 at 7:49




          $begingroup$
          You can read the Wikipedia article I just added to my answer. I took an example directly from there.
          $endgroup$
          – William Sun
          Dec 7 '18 at 7:49


















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