Show that \(\sum\limits^{\infty}_{k = 0} k^2 x^k = x(1+x)/(1-x)^3\) for \(\lvert x \rvert < 1\).

Based on equation A.8 we know that

\[\sum\limits^{\infty}_{k=0} kx^k = \frac{x}{(1-x)^2}\]

for \(\lvert x \rvert\). This isn’t quite the equation we want, so let’s differentiate each side of it:

\[\begin{split} \sum\limits^{\infty}_{k=0} k \cdot kx^{k-1} &= \frac{(1-x)^2 + 2x(1-x)}{(1-x)^4} \\ &= \frac{(1-x) + 2x}{(1-x)^3} \\ &= \frac{1+x}{(1-x)^3} \\ \end{split}\]

Finally, we multiply both sides by \(x\) to reach the desired result:

\[\sum\limits^{\infty}_{k = 0} k^2 x^k = \frac{x(1+x)}{(1-x)^3}\]