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

Question

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.

Answer 1

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.

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

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.

Answer 2

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

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

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