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

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"?

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.
Note that for large $$n$$, $$H_n\sim \ln n$$, so $$\ln n\approx 360$$ means that $$n\approx 2.2\times 10^{156}$$.
• It is misleading to give so many significant figures! The approximation $H_n\sim \ln n$ is not nearly so precise. – TonyK Jan 22 at 14:51