Show that $$\sum_{k = 0}^{\infty} (k - 1)/2^k = 0$$.

From chapter text $$\sum_{k = 0}^{\infty} x^k = \frac 1 {(1 - x)}$$ (Eqn. A.6) and $$\sum_{k = 0}^{\infty} kx^k = \frac x {(1 - x)^2}$$ (Eqn. A.8).

So, we can write:

\begin {align} & \sum_{k = 0}^{\infty} kx^k - \sum_{k = 0}^{\infty} x^k \tag {1} \\ = & \frac x {(1 - x)^2} - \frac 1 {(1 - x)} \\ = & \frac {x - (1 - x)} {(1 - x)^2} \\ = & \frac {2x - 1} {(1 - x)^2} \end {align}

Now, we can rearrange the expression in question as follows:

\begin {align} \sum_{k = 0}^{\infty} \frac {k - 1} {2^k} & = \sum_{k = 0}^{\infty} \frac k {2^k} - \sum_{k = 0}^{\infty} \frac 1 {2^k} \\ & = \sum_{k = 0}^{\infty} k \cdot \left(\frac 1 2\right)^k - \sum_{k = 0}^{\infty} \left(\frac 1 2\right)^k \end {align}

Note that this rearranged version is nothing but (1) with $$x = \frac 1 2$$.

$\therefore \sum_{k = 0}^{\infty} \frac {k - 1} {2^k} = \frac {2 \cdot \frac 1 2 - 1} {\left(1 - \frac 1 2\right)^2} = 0$