This question relates to proving a random variable N on $(\mathbf{\Omega },\mathfrak{F},\mathbb{P})$ which has image {0,1,2,...} has expectation $\sum_{k=0}^{\infty} \mathbb{P}(N> k)$ if the sum exists.
So far my proof is as follows, let $f$ and $F$ be the probability mass function and cumulative distribution function respectively.
The expectation of N is $$\begin{align}\sum_{k=\left \{0,1,2,... \right \}} k\mathbb{P}(N=k) &=\lim_{n\rightarrow \infty}\sum_{k=0}^{n} k\mathbb{P}(N=k) \\&=\lim_{n\rightarrow \infty}\sum_{k=0}^{n} kf(k) \\&=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} k[F(k)-F(k-1)] \text{ (k=0 term is 0)}\end{align}$$ The sum simplifies to: $$\begin{align} \lim_{n\rightarrow \infty}\sum_{k=1}^{n} (F(n)-F(k-1)) &=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \mathbb{P}(k-1<N<n) \\&=\lim_{n\rightarrow \infty}\sum_{k=0}^{n} \mathbb{P}(k<N<n)\end{align}$$
My question is as follows: in my final line, does the nature of the analysis of the sum allow me to take the limit within it so it becomes $\sum_{k=0}^{\infty} \lim_{n\rightarrow \infty}\mathbb{P}(k<N<n)$, which is indeed $\sum_{k=0}^{\infty} \mathbb{P}(N> k)$ the required result? Or is there a more approproate alternative route that I can take?
Yes, it is legitimate to take the limit inside. You can apply Monotone Convergence Theorem noting that $I_{\{k \leq n\}}I_{\{N<n\}}$ is non-negative and increasing.