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