0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ No chance for an exact formula. There are very sharp estimates though. $\endgroup$ – A. Pongrácz Jan 2 at 23:47
  • $\begingroup$ Also, this looks like a duplicate. See here $\endgroup$ – jmerry Jan 2 at 23:48
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $10.00004300827580769435$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.