This question already has an answer here:

We know that $\sum_{k=1}^{\infty}\frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots$ diverges.

But for any natural number $n$, $\sum_{k=1}^{n}\frac{1}{k}$ is finite.

The question is; how to compute $\sum_{k=1}^{n}\frac{1}{k}$ for sum natural number $n$, explicitly?

Say $n=50$, how to compute $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{49}+\frac{1}{50}$, explicitly?


marked as duplicate by Martin R, Masacroso, Arjang, Scientifica, José Carlos Santos Oct 30 '18 at 10:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


There is no explicit or exact value but accurate asymptotics $$\sum_{k=1}^n\frac 1k=H_n$$ and, for large values of $n$ $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}-\frac{1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ For $n=50$, the exact value would be $$H_{50}=\frac{13943237577224054960759}{3099044504245996706400}\approx 4.499205338$$ while the above approximation would give $$\gamma +\log (50)+\frac{299}{30000}\approx 4.499205337$$ which is not too bad (I hope !).


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