Let $X$ have a geometric distribution such that $p_{X}(x) = \textbf{P}(X = x) = q^{x−1}p$, where $x \geq 1$. Show that $\textbf{E}(X^{-1}) = \log(p^{1/p−1})$.
MY ATTEMPT
According to the properties of the expectation, we have \begin{align*} \textbf{E}(g(X)) = \sum_{x=1}^{\infty}g(x)p_{X}(x) \end{align*}
In the particular case when $g(X) = X^{-1}$, the problem reduces to determine \begin{align*} \textbf{E}(X^{-1}) = \sum_{x=1}^{\infty}\frac{q^{x-1}p}{x} = p\sum_{x=1}^{\infty}\frac{q^{x-1}}{x} \end{align*}
Now consider the series
\begin{align*} f(w) & = 1 + \frac{w}{2} + \frac{w^{2}}{3} + \ldots \Rightarrow F(w) = \int_{0}^{w}f(u)\mathrm{d}u = \frac{w}{1^{2}} + \frac{w^{2}}{2^{2}} + \frac{w^{3}}{3^{2}} + \ldots \end{align*}
If I am on the right track, $\textbf{E}(X^{-1}) = pF^{\prime}(1-p)$. However, I am unable to find out the closed formula for both $f$ and $F$. Am I failing at some point? Otherwise, could someone help me out? Thanks!
EDIT
According to Alex's contribution, we can rewrite the given expression as \begin{align*} \textbf{E}(X^{-1}) = \frac{p}{q}\sum_{x=1}^{\infty}\frac{q^{x}}{x} \end{align*}
Therefore, if we conveniently define \begin{align*} G(w) & = w + \frac{w^{2}}{2} + \frac{w^{3}}{3} + \ldots \Rightarrow G^{\prime}(w) = g(w) = 1 + w + w^{2} + \ldots = \frac{1}{1-w} \Rightarrow\\\\ G(w) & = \log(1-w) \Rightarrow \textbf{E}(X^{-1}) = \frac{p}{q}G(q) = \frac{p}{1-p}\log(1 - (1 - p)) = \frac{p}{1-p}\log(p) \end{align*}
Am I missing something?