$\sum \frac{1}{k}$ , sum of the harmonic series How do we sum up this series in terms of $n$?
$$1 + \frac{1}{2} + \ldots+\frac{1}{n}$$
Can we create a formula in terms of $n$ for this series?
 A: $$\sum_{k=1}^n\frac{1}{k}=\ln n+\gamma+\Theta\big(\frac{1}{2n}\big)$$ where $\gamma $ is Euler–Mascheroni constant.
A: As I explained here, for $n>1$,
$$1/1+1/2+\cdots+1/n =\log(n+1/2) + \gamma + \varepsilon(n)$$
with
$$0 < \varepsilon(n) < 1/(24n^2)$$
which gives a good approximation.
A: Let $\displaystyle H_n=\sum_{k=1}^n\frac{1}{k}$.
First, we'll prove that the sequence $(u_n)=(H_n-\log n)$ converges by showing that the series $\displaystyle\sum_n u_{n}-u_{n-1}$ is convergent. We have:
$$u_{n}-u_{n-1}=\frac{1}{n}+\log(1-\frac{1}{n})\sim-\frac{1}{2n^2},$$
so the series $\displaystyle\sum_n u_n$ is convergent by comparaison with the Riemann series.
If we denote by $\gamma$ the limit of the sequence $(u_n)$ then we have the formula:
$$\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+o(1).$$
Moreover, we have:
$$\sum_{k=n+1}^{\infty}u_{k}-u_{k-1}=\gamma-u_n\sim-\sum_{k=n+1}^{\infty}\frac{1}{2k^2}\sim-\frac{1}{2n},$$ 
hence, we have an improved formula:
$$\sum_{k=1}^n\frac{1}{k}=\log(n)+\gamma+\frac{1}{2n}+o(\frac{1}{n}).$$
