Show that $$\sum_{k = 0}^{\infty} k^2x^k = x(1 + x)/(1 - x)^3$$ for $$0 < \vert x \vert < 1$$.

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

By differentiating this with respect to $$x$$, we get:

\begin {align} \sum_{k = 0}^{\infty} k^2x^{k - 1} & = \frac 1 {(1 - x)^2} + x \cdot \left( -\frac {2x - 2} {(1 - x)^4}\right) \\ & = \frac 1 {(1 - x)^2} + x \cdot \left( \frac 2 {(1 - x)^3}\right) \\ & = \frac {(1 - x) + 2x } {(1 - x)^3} \\ & = \frac {1 + x} {(1 - x)^3} \end {align}

Multiplying both sides by $$x$$, we get:

$\sum_{k = 0}^{\infty} k^2x^k = \frac {x(1 + x)} {(1 - x)^3}$