Is there any closed form for the finite sum $1+\frac12+\frac13+\frac14+\dots+\frac1n?$

Question

I know that the infinite summation $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}+...$$ is divergent and also the sequence $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}-\ln n$$ is convergent.

I wonder that whether there exist a formula (closed form) for the obtain value of the following finite summation. $$1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}$$ If not why?

Answer 1

$$H_n~=~\psi(n+1)+\gamma~=~-\int_0^1\ln\Big(1-\sqrt[n]x\Big)~dx~=~\int_0^1\frac{1-x^n}{1-x~~}~dx,$$ where $\psi$ is the digamma function, and $\gamma$ is the Euler-Mascheroni constant. See harmonic numbers for more details.

Answer 2

$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}=\sum_{i=1}^{n} \frac{1}{i}=H_n$$

where $H_n$ is the $n^{th}$ harmonic number.