Answer
See explanation
Work Step by Step
Let $p_X(n),\ n= 0, 1, 2, 3, \ldots,$ be the pdf of $X$.
$\begin{align*}
E(X) &= \sum_{-\infty}^\infty n\cdot p_X(n) & [\ \text{Definition of}\ E(X)\ ] \\
&= \sum_{n=0}^\infty n\cdot p_X(n) & [\ \text{since}\ p_X(n) = 0, n \in \mathbb{Z}^-\ ] \\
&= \sum_{n=1}^\infty n\cdot p_X(n) & [\ \text{since}\ 0\cdot p_X(0) = 0\ ] \\
&= \sum_{n=1}^\infty \overbrace{\left(\sum_{k=1}^n 1\right)}^{=\; n} \cdot p_X(n) & \left[\ \text{since}\ \sum_{k=1}^n 1 = n\ \right] \\
&= \sum_{n=1}^\infty \sum_{k=1}^n p_X(n) \\
& \qquad\qquad [\ \text{for each fixed}\ n,\ n=1,2,3,\ldots, \\
& \qquad\qquad k\ \text{goes from}\ 1\ \text{to}\ n\ \text{(upwards; see diagram)}\ ] \\
&= \sum_{k=1}^\infty \sum_{n=k}^\infty p_X(n) \\
& \qquad\qquad \color{green}{[\ \text{switch order of summation}\colon}\ \\
& \qquad\qquad \color{green}{\text{for each fixed}\ k,\ n=1,2,3,\ldots,} \\
& \qquad\qquad \color{green}{n\ \text{goes from}\ k\ \text{to}\ \infty\ \text{(rightwards; see diagram)}\ ] }\\
E(X) &= \sum_{k=1}^\infty P(X\ge k) \\
\end{align*}$
