Show that \(\sum\limits^{\infty}_{k=0} \frac{(k-1)}{2^k} = 0\).

We can use formulas A.6 and A.8 to deconstruct this summation:

\[\begin{split} \sum\limits^{\infty}_{k=0} \frac{(k-1)}{2^k} &= \sum\limits^{\infty}_{k=0} \frac{k}{2^k} - \sum\limits^{\infty}_{k=0} \frac{1}{2^k} \\ &= \sum\limits^{\infty}_{k=0} \frac{k}{2^k} - \frac{1}{1 - \frac{1}{2}} \\ &= \sum\limits^{\infty}_{k=0} k \left( \frac{1}{2} \right)^k - 2 \\ &= \sum\limits^{\infty}_{k=0} \frac{\frac{1}{2}}{\left(1 - \frac{1}{2} \right)^2} - 2 \\ &= \frac{1/2}{1/4} - 2 \\ &= 2 - 2 \\ &= 0 \\ \end{split}\]