\begin{equation*}
\sum_{k=2}^{n}\frac{1}{k}\le\ln n
\end{equation*}
To see this, look at the curve \(1/x\), and integrate it from 1 to \(n\). This will give you the area under the curve, but the “approximation” to the curve will be given by the sum:
\begin{equation*}
1+x\le e^{x}
\end{equation*}