Derivation of E(X) for a geometric distribution I'm having trouble following the derivation of the expected value for the geometric distribution.
I've reached:
$$\sum_{k=1}^{\infty}
    p \cdot k(1-p)^{k-1} = 
    p(1 + 2(1-p) + 3(1-p)^2 + ...)
$$
The next step rewrites $k$ as an infinite sum of ascending values, that is:
$$=p( \sum_{k=1}^{\infty}\ (1-p)^{k-1} + \sum_{k=2}^{\infty}\ (1-p)^{k-1} + \sum_{k=3}^{\infty}\ (1-p)^{k-1} + ...)$$
Why exactly is it that we can represent $k$ as this infinite sum?
 A: Consider the following double infinite sum:
\begin{array}{cccccc}
p & +p(1-p) &  +p(1-p)^2 & + p(1-p)^3 & + p(1-p)^4 & + \cdots \\
  & +p(1-p) &  +p(1-p)^2 & + p(1-p)^3 & + p(1-p)^4 & + \cdots \\
  &         &  +p(1-p)^2 & + p(1-p)^3 & + p(1-p)^4 & + \cdots \\
  &         &            & + p(1-p)^3 & + p(1-p)^4 & + \cdots \\
  &         &            &            & + p(1-p)^4 & + \cdots \\
  &         &            &            &            & + \cdots \\
\end{array}
If we first sum the columns, then the $k^{\text{th}}$ column sums to $k p(1-p)^{k-1}$, and we get the sum you're starting with.
If we first sum the rows, then the $j^{\text{th}}$ row sums to $p \displaystyle\sum_{k=j}^{\infty} (1-p)^{k-1}$, and we get the second sum.

Another way to see this: in the expression $$p\left( \sum_{k=1}^{\infty}\ (1-p)^{k-1} + \sum_{k=2}^{\infty}\ (1-p)^{k-1} + \sum_{k=3}^{\infty}\ (1-p)^{k-1} + \dots\right),$$ the term $p(1-p)^{j-1}$ appears in exactly $j$ different sums.
A: We have :
$$\begin{array}{lcl}
\displaystyle \sum_{k=1}^{+\infty} p k (1 - p)^{k - 1} & = & \displaystyle \sum_{k=1}^{+\infty} p \left(\sum_{i = 1}^k 1\right) (1 - p)^{k - 1} \\[3mm]
& = & \displaystyle \sum_{k=1}^{+\infty} \sum_{i = 1}^k p (1 - p)^{k - 1} \\[3mm]
& = & \displaystyle \sum_{i = 1}^{+\infty} p \sum_{k=i}^{+\infty} (1 - p)^{k - 1} \\[3mm]
& = & \displaystyle p \sum_{k=1}^{+\infty} (1 - p)^{k - 1} + p \sum_{k=2}^{+\infty} (1 - p)^{k - 1} + p \sum_{k=3}^{+\infty} (1 - p)^{k - 1} + \cdots
\end{array}$$
