3rd Factorial Moment of X ~ geo(p) I'm working through Pitman's Probability (1993) problem 23b, page 221 (for knawledge not school). 
Specifically, I am having trouble calculating 
$E[(X)_3] = G^{(3)}(1)$, with $G^{(k)}(z) = \sum_{i=k}^{\infty}(P(X=i)\cdot(i)_k\cdot z^{i-k}$), where $G(z)$ is the probability generating function of $X$ distributed geometric(p); i.e. number of trials for first success. Pitman claims $E[(X)_k] = G^{(k)}(1)$
Pitman and WolframAlpha give $E[(X)_k] = 6(q/p)^3$. My solution gives $6q^2/p^3$.
Here's my process: 
$$ \begin{align} G^{(3)}(1) &= q^2p(3)_3 + q^3p(4)_3 + q^4p (5)_3 + \cdots \\ 
&= q^2p \cdot \Sigma_1,\quad\Sigma_1 = 3\cdot2\cdot 1 \ + 4\cdot 3\cdot 2q\ +5\cdot 4\cdot 3q^2 +\cdots\\
\end{align} $$
Solving for $\Sigma_1$, we do the subtraction trick:
$$\begin{align}\Sigma_1 - q\Sigma_1 &= 3\cdot2\cdot (1 + 3q + 6q^2 + 10q^3 + 15q^4+\cdots)\\
&=6\cdot \Sigma_2, \quad \Sigma_2 = 1 + 3q + 6q^2 + 10q^3 + 15q^4+\cdots
\end{align}
$$
Solving for $\Sigma_2$, we do the old subtraction trick again:
$$\begin{align}\Sigma_2 - q\Sigma_2 &= 1 + 2q+ 3q^2 +4q^3+\cdots = \Sigma_3\\
\end{align}
$$
Solving for $\Sigma_3$, we do grandpa's subtraction trick again to get: $$\Sigma_3  = 1/(1-q)^2 \Rightarrow \Sigma_2 = 1/(1-q)^3 \Rightarrow \Sigma_1 = 6/(1-q)^4
$$ 
Hence $E[(X)_3] = q^2 p \cdot \frac{6}{(1-q)^4} = 6q^2/p^3$.
Did my subtractions fail? Are my assumptions wrong? What happened?
Edit: Pitman mentions in problem 23a) that $X$ is distributed $geometric(p)$ on ${0,1,2,...}$, which he explains in problem 6 is equivalent to the standard geometric distribution. I did not do problem 6 and used the wrong geometric distribution. Sad. The clever posters point out the difference more explicitly.

In general, if $P(z)$ is the generating function of a random variable $X$ with probability mass function $\{p_k\}$, then since $0\leqslant p_k\leqslant 1$ and $\sum_{k=1}^\infty p_k=1$, it follows from dominated convergence that
$$
\frac{\mathsf d}{\mathsf dz} P(z) = \frac{\mathsf d}{\mathsf dz} \sum_{k=1}^\infty p_kz^k =  \sum_{k=1}^\infty kp_kz^{k-1},\quad 0<z<1.
$$
By induction we see that 
$$
\frac{\mathsf d^n}{\mathsf dz^n} P(z) = \sum_{k=n}^\infty (k)_n p_k z^{k-n},\quad n\geqslant1
$$
where $(k)_n = \frac{k!}{(k-n)!}$ denotes the falling factorial. Monotone convergence then yields
$$
\lim_{z\uparrow 1}\frac{\mathsf d^n}{\mathsf dz^n} P(z) = \sum_{k=0}^\infty (k)_np_k = \mathbb E[(X)_k].
$$
Here $$\mathbb E[z^X] := G(z) = \sum_{k=1}^\infty kp(1-p)^{k-1}z^k $$
so 
\begin{align}
\mathbb E[(X)_3] &= \lim_{z\uparrow1} G'(z)\\
&= \lim_{z\uparrow1} p(1-p)^2\sum_{k=3}^\infty k(k-1)(k-2)((1-p)z)^{k-3}\\
&= p(1-p)^2\sum_{k=0}^\infty (k+1)(k+2)(k+3)\\
&= p(1-p)^2\left[\frac{\mathsf d^3}{\mathsf dp^3}\sum_{k=0}^\infty (1-p)^k \right]\\
&= p(1-p)^2 \frac{\mathsf d^3}{\mathsf dp^3} \left(\frac 1p \right)\\
&= p(1-p)^2 \cdot \frac 6{p^4}\\
&= \frac{6(1-p)^2}{p^3}.
\end{align}
Your solution is indeed correct.
