\displaystyle $\begin{align*}
\sum_{k = 0}^{\infty} x^k & = \frac{1}{1 - x} & / \int \\
\sum_{k = 0}^{\infty} \frac{x^{k + 1}}{k + 1} & = -\ln(1 - x) & / : x \\
\sum_{k = 0}^{\infty} \frac{x^k}{k + 1} & = -\frac{\ln(1 - x)}{x} & / x = 1 - p \\
\sum_{k = 0}^{\infty} \frac{(1 - p)^k}{k + 1} & = -\frac{\ln p}{1 - p}
\end{align*}$