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 \(x = \frac 1 2\).
\[\therefore \sum_{k = 0}^{\infty} \frac {k - 1} {2^k} = \frac {2 \cdot \frac 1 2 - 1} {\left(1 - \frac 1 2\right)^2} = 0\]