<s>[INST] Third Moment of Geometric Series? $\sum_{x=1}^{\infty}x(1-p)^{x-1} = \frac{1}{p^2}$
$\sum_{x=1}^{\infty}x^2(1-p)^{x-1} = \frac{2-p}{p^3}$
$\sum_{x=1}^{\infty}x^3(1-p)^{x-1} =$ ... ?
The textbook gives the first 2 moments, but not the third one. I could not find the forumla on Google as well. I think it'll be useful in case I need to find $E(X^3)$ where $X$ is a geometric distribution.
 [/INST] Using summation of series.
$$S=\sum_{k=1}^\infty x^3(1-p)^{k-1}\\ 
  -(1-p)S=-\sum_{k=1}^\infty (x-1)^3(1-p)^{k-1}$$
We add these two and get 
$$pS=\sum_{k=1}^\infty (x^3-(x-1)^3)(1-p)^{1-k} \\ =\sum_{k=1}^\infty (x^2+x(x-1)+(x-1)^2)(1-p)^{1-k} \\ =\sum_{k=1}^\infty x^2(1-p)^k+\sum_{k=1}^\infty (x-1)^2(1-p)^{1-k}+\sum_{k=1}^\infty x(x-1)(1-p)^{1-k}\\ 
=E(X^2)+(1-p)E(X^2) + K$$
$$K=\sum_{k=1}^\infty x(x-1)(1-p)^{1-k}=\sum_{k=1}^\infty x(x+1)(1-p)^{k} \\ 
(1-p)K=\sum_{k=1}^\infty x(x-1)(1-p)^k \\ K-(1-p)K=pK=\sum_{k=1}^\infty (x(x+1)-x(x-1))(1-p)^{k} \\ pK=\sum_{k=1}^\infty 2x(1-p)^{k}=2(1-p)E(X)$$
Therefore $$pS=E(X^2)+(1-p)E(X)^2+\frac{2(1-p)}{p}E(X) \\ = \frac{(2-p)^2}{p^3}+\frac{2(1-p)}{p^3}=\frac{p^2-6p+6}{p^3}$$

$$S=\frac{p^2-6p+6}{p^4}$$

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