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:

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

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