How many terms are required for harmonic series of degrees to “cover” a full $360^text{o}$ circle?












2












$begingroup$


This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
$$
H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
$$



Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
$$
S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
$$



Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
$$
H_n^text{o} ge 360^text{o}
$$



Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:





  1. Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?

  2. If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?







calculus sequences-and-series summation harmonic-numbers







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
    $$
    H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
    $$



    Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
    $$
    S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
    $$



    Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
    $$
    H_n^text{o} ge 360^text{o}
    $$



    Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:





    1. Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?

    2. If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?







    calculus sequences-and-series summation harmonic-numbers







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
      $$
      H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
      $$



      Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
      $$
      S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
      $$



      Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
      $$
      H_n^text{o} ge 360^text{o}
      $$



      Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:





      1. Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?

      2. If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?







      calculus sequences-and-series summation harmonic-numbers







      share|cite|improve this question











      $endgroup$




      This question accidentally came to my mind when reading about harmonic series. I've never been able to find an answer on the Internet. Consider $H_n$ which is the $n$-th harmonic number:
      $$
      H_n = 1 + {1over 2} + {1over 3} + cdots + {1over n}
      $$



      Not sure whether it's valid but suppose we have a circle which's full rotation is known to be $360^text{o}$. Lets define the following series:
      $$
      S_n^text{o} = (1)^text{o} + left({1over 2}right)^text{o} + left({1over 3}right)^text{o} + cdots
      $$



      Since harmonic series is divergent and its sum tends to infinity, then at some point we should have "covered" the whole circumference. If we now define an "$n$-th harmonic degree number" as $H_n^text{o}$. Then we have to solve the following inequality for $n$:
      $$
      H_n^text{o} ge 360^text{o}
      $$



      Please note that i'm not very familiar with series and only have basic calculus knowledge like limits, derivatives and Taylor expansion. Also I may have misused a lot of terms in this question, so please comment in order for me to improve it. Apart from that I'm basically interested in two things:





      1. Is it valid to consider the harmonic sum of degrees rather than the sum of rationals?

      2. If so what would be the way to find the index of $H_n^text{o}$ such that the whole circle is "covered"?







      calculus sequences-and-series summation harmonic-numbers




      calculus 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 22 at 14:50









      José Carlos Santos

      164k22131234




      164k22131234










      asked Jan 22 at 14:35









      romanroman

      2,30921224




      2,30921224






















          2 Answers
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          3












          $begingroup$

          Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Indeed. Do you mind if I edit my answer, adding to it your suggestion?
            $endgroup$
            – José Carlos Santos
            Jan 22 at 14:49










          • $begingroup$
            whoa, that's a long way round. Thanks!
            $endgroup$
            – roman
            Jan 22 at 14:53



















          3












          $begingroup$

          Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
            $endgroup$
            – TonyK
            Jan 22 at 14:51











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Indeed. Do you mind if I edit my answer, adding to it your suggestion?
            $endgroup$
            – José Carlos Santos
            Jan 22 at 14:49










          • $begingroup$
            whoa, that's a long way round. Thanks!
            $endgroup$
            – roman
            Jan 22 at 14:53
















          3












          $begingroup$

          Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Indeed. Do you mind if I edit my answer, adding to it your suggestion?
            $endgroup$
            – José Carlos Santos
            Jan 22 at 14:49










          • $begingroup$
            whoa, that's a long way round. Thanks!
            $endgroup$
            – roman
            Jan 22 at 14:53














          3












          3








          3





          $begingroup$

          Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.






          share|cite|improve this answer











          $endgroup$



          Since $H_n$ is close to $log n$ if $n$ is large enough, you will have to take $n$ about $e^{360-gamma}$, which is about $1.25times10^{156}$. Here, $gamma$ is the Euler-Mascheroni constant.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 22 at 14:53

























          answered Jan 22 at 14:44









          José Carlos SantosJosé Carlos Santos

          164k22131234




          164k22131234












          • $begingroup$
            Indeed. Do you mind if I edit my answer, adding to it your suggestion?
            $endgroup$
            – José Carlos Santos
            Jan 22 at 14:49










          • $begingroup$
            whoa, that's a long way round. Thanks!
            $endgroup$
            – roman
            Jan 22 at 14:53


















          • $begingroup$
            Indeed. Do you mind if I edit my answer, adding to it your suggestion?
            $endgroup$
            – José Carlos Santos
            Jan 22 at 14:49










          • $begingroup$
            whoa, that's a long way round. Thanks!
            $endgroup$
            – roman
            Jan 22 at 14:53
















          $begingroup$
          Indeed. Do you mind if I edit my answer, adding to it your suggestion?
          $endgroup$
          – José Carlos Santos
          Jan 22 at 14:49




          $begingroup$
          Indeed. Do you mind if I edit my answer, adding to it your suggestion?
          $endgroup$
          – José Carlos Santos
          Jan 22 at 14:49












          $begingroup$
          whoa, that's a long way round. Thanks!
          $endgroup$
          – roman
          Jan 22 at 14:53




          $begingroup$
          whoa, that's a long way round. Thanks!
          $endgroup$
          – roman
          Jan 22 at 14:53











          3












          $begingroup$

          Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
            $endgroup$
            – TonyK
            Jan 22 at 14:51
















          3












          $begingroup$

          Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
            $endgroup$
            – TonyK
            Jan 22 at 14:51














          3












          3








          3





          $begingroup$

          Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.






          share|cite|improve this answer











          $endgroup$



          Note that for large $n$, $H_nsim ln n$, so $ln napprox 360$ means that $napprox 2.2times 10^{156}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 22 at 14:56

























          answered Jan 22 at 14:44









          rogerlrogerl

          18k22747




          18k22747








          • 1




            $begingroup$
            It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
            $endgroup$
            – TonyK
            Jan 22 at 14:51














          • 1




            $begingroup$
            It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
            $endgroup$
            – TonyK
            Jan 22 at 14:51








          1




          1




          $begingroup$
          It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
          $endgroup$
          – TonyK
          Jan 22 at 14:51




          $begingroup$
          It is misleading to give so many significant figures! The approximation $H_nsim ln n$ is not nearly so precise.
          $endgroup$
          – TonyK
          Jan 22 at 14:51


















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