# Sum to n terms of the harmonic series [duplicate]

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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 SantosOct 30 '18 at 10:31

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## 1 Answer

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 !).