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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: $$ H_n = 1 + {1\over 2} + {1\over 3} + \cdots + {1\over n} $$

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: $$ S_n^\text{o} = (1)^\text{o} + \left({1\over 2}\right)^\text{o} + \left({1\over 3}\right)^\text{o} + \cdots $$

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"?
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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 $1.25\times10^{156}$. Here, $\gamma$ is the Euler-Mascheroni constant.

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  • $\begingroup$ 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
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Note that for large $n$, $H_n\sim \ln n$, so $\ln n\approx 360$ means that $n\approx 2.2\times 10^{156}$.

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

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