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

Question

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"?

Answer 1

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.

Answer 2

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