Is there a way to sum (part of) the harmonic series to a given total?












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For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.






sequences-and-series summation harmonic-numbers







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  • $begingroup$
    No chance for an exact formula. There are very sharp estimates though.
    $endgroup$
    – A. Pongrácz
    Jan 2 at 23:47










  • $begingroup$
    Also, this looks like a duplicate. See here
    $endgroup$
    – jmerry
    Jan 2 at 23:48
















0












$begingroup$


For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.






sequences-and-series summation harmonic-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    No chance for an exact formula. There are very sharp estimates though.
    $endgroup$
    – A. Pongrácz
    Jan 2 at 23:47










  • $begingroup$
    Also, this looks like a duplicate. See here
    $endgroup$
    – jmerry
    Jan 2 at 23:48














0












0








0





$begingroup$


For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.






sequences-and-series summation harmonic-numbers







share|cite|improve this question











$endgroup$




For example, $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6$ can be solved by traditional arithmetic of $1 + 360/720 + 240/720...$ etc. to get a total of $2.45$, but if I wanted the total to say, $1/12367$, the value of which sum is about $10$, is there a fast process for acquiring this total? I'm hoping there's something similar to Gauss's solution to arithmetic sequences.






sequences-and-series summation harmonic-numbers




sequences-and-series summation harmonic-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 0:12









Eevee Trainer

7,04321337




7,04321337










asked Jan 2 at 23:41









TristianTristian

1769




1769












  • $begingroup$
    No chance for an exact formula. There are very sharp estimates though.
    $endgroup$
    – A. Pongrácz
    Jan 2 at 23:47










  • $begingroup$
    Also, this looks like a duplicate. See here
    $endgroup$
    – jmerry
    Jan 2 at 23:48


















  • $begingroup$
    No chance for an exact formula. There are very sharp estimates though.
    $endgroup$
    – A. Pongrácz
    Jan 2 at 23:47










  • $begingroup$
    Also, this looks like a duplicate. See here
    $endgroup$
    – jmerry
    Jan 2 at 23:48
















$begingroup$
No chance for an exact formula. There are very sharp estimates though.
$endgroup$
– A. Pongrácz
Jan 2 at 23:47




$begingroup$
No chance for an exact formula. There are very sharp estimates though.
$endgroup$
– A. Pongrácz
Jan 2 at 23:47












$begingroup$
Also, this looks like a duplicate. See here
$endgroup$
– jmerry
Jan 2 at 23:48




$begingroup$
Also, this looks like a duplicate. See here
$endgroup$
– jmerry
Jan 2 at 23:48










2 Answers
2






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oldest

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2












$begingroup$

There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$



$epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.



$gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.



$${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$



Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.



    If you want more accuracy, you could use
    $$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
    n^2}+Oleft(frac{1}{n^4}right)$$



    Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






      active

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      $begingroup$

      There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$



      $epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.



      $gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.



      $${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$



      Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$



        $epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.



        $gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.



        $${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$



        Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$



          $epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.



          $gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.



          $${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$



          Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.






          share|cite|improve this answer









          $endgroup$



          There is a sort of way to sum the series, but it's not exact. $$1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + ... + frac{1}{n} = sum_{k=1}^n frac{1}{k} = ln(n) + gamma + epsilon_n$$



          $epsilon_n$ is an error constant proportional to $1/2n$, and thus $epsilon_n to 0$ as $n to infty$.



          $gamma$ denotes the Euler-Mascheroni constant, defined by the limiting difference between the natural logarithm and the harmonic series, i.e.



          $${displaystyle {begin{aligned}gamma &=lim _{nto infty }left(-ln n+sum _{k=1}^{n}{frac {1}{k}}right)\[5px]end{aligned}}} approx 0.5772156649$$



          Thus, you can approximate the sum of the first $n$ natural reciprocals by $ln(n)$, plus $gamma$ if you so choose, with greater accuracy as $n to infty$. However, there is no known exact formula for these partial sums.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 0:09









          Eevee TrainerEevee Trainer

          7,04321337




          7,04321337























              2












              $begingroup$

              Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.



              If you want more accuracy, you could use
              $$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
              n^2}+Oleft(frac{1}{n^4}right)$$



              Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.



                If you want more accuracy, you could use
                $$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
                n^2}+Oleft(frac{1}{n^4}right)$$



                Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.



                  If you want more accuracy, you could use
                  $$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
                  n^2}+Oleft(frac{1}{n^4}right)$$



                  Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$






                  share|cite|improve this answer









                  $endgroup$



                  Eevee Trainer gave the right idea that is to say considering the asymptotics of harmonic numbers.



                  If you want more accuracy, you could use
                  $$ sum_{k=1}^n frac{1}{k}=H_n=gamma +log left({n}right)+frac{1}{2 n}-frac{1}{12
                  n^2}+Oleft(frac{1}{n^4}right)$$



                  Trying with $n=12367$, the "exact" value is $10.00004300827580769471$ while the above approximation would give $10.00004300827580769435$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 9:42









                  Claude LeiboviciClaude Leibovici

                  123k1157134




                  123k1157134






























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