29.51. Calculate $\lim _{n \rightarrow \infty} \sum_{k=n}^{2 n} \frac{1}{k}$.
29.51. Answer: $\ln 2$. Note that the considered sum is equal to

$$
\frac{1}{n}\left(1+\frac{1}{1+1 / n}+\frac{1}{1+2 / n}+\ldots+\frac{1}{1+n / n}\right) \rightarrow \int_{1}^{2} \frac{d x}{1+x}=\ln 2
$$

Remark. Another proof can be found in the solution to problem 30.8 a).