Durch die Untersumme erkennt man
\(\begin{aligned} \sum \limits_{k=1}^{n} \frac{1}{k} \geqslant \int \limits_{1}^{n} \frac{1}{x+1} d x\end{aligned} \)
und somit
\( \begin{aligned} \sum \limits_{k=0}^{n^{2}-n} \frac{1}{n+k}=\sum \limits_{k=1}^{n^{2}} \frac{1}{k}-\sum \limits_{k=1}^{n-1} \frac{1}{k} & \geqslant \int \limits_{1}^{n^{2}} \frac{1}{x+1} d x-\int \limits_{1}^{n-1} \frac{1}{x+1} d x \\ &=\int \limits_{n-1}^{n^{2}} \frac{1}{x+1} d x \\ &=[\ln (x+1)]_{n-1}^{n^{2}} \\ &=\ln \left(n^{2}+1\right)-\ln (n) \geqslant \ln (n) \end{aligned} \)
