This is a fundamental stochastic process useful in modelling arrivals and interarrival times. Some definitions will make the usage clear.
Let $S_{i}$ denote the $i$th renewal time or the time when the $i$th arrival takes place. By definition, $S_{0} = 0$. We can also define \begin{align} S_{n} &= S_{n-1} + \xi_{n}\newline S_{n} &= \xi_{1} + \xi_{2} + \cdots + \xi_{n-1} \end{align} where $\xi_{i}$ are positive ($P(\xi > 0) = 1$) independent identically distributed variables representing the interarrival times. We also define \begin{align} N_{t} &= \argmax_{k} { S_{k} \leq t }\newline { S_{n} > t } &= { N_{t} < n } \end{align} or, $N_{t}$ is simply the number of arrivals till the time $t$.
Define the following quantity \begin{align} F^{n*} &= F_{\xi} * \ldots * F_{\xi} \text{ $n$ times}\newline u(t) &= \sum_{i=1}^{\infty} F^{n*}(t) \end{align} It can be shown that the function $u(t)$ converges. The expectation of $N_{t}$ then becomes \begin{align} E[N_{t}] &= E[\text{number of $n$ such that $S_{n} \leq t$}]\newline &= E[\sum_{n=1}^{\infty} I(S_{n} \leq t)] \quad \text{ sum of Indicators will equal $n$}\newline &= \sum_{n=1}^{\infty} P(S_{n} \leq t) \quad \text{ since $E[$Indicator$]$ is just the function inside indicator}\newline &= \sum_{n=1}^{\infty} F^{n*}(t) \quad \text{ by defining cumulative as sum of $\xi$s}\newline &= u(t) \end{align}
For a density function $f$ defined from $\mathbb{R}^{\geq 0} \to \mathbb{R}$, Laplace transform is \begin{align} L_{f}(s) = \int_{\mathbb{R}^{\geq 0}} e^{-sx} f(x) dx \end{align} The following properties hold for this transform
If $f$ is a probability density function, then \begin{align} E[e^{-sx}] = L_{f}(s) \end{align}
if $f_{1}$ and $f_{2}$ are two probability density functions, then \begin{align} L_{f_{1}*f_{2}}(s) = L_{f_{1}}(s) L_{f_{2}}(s) \end{align}
If $F$ is the cumulative probability distribution for $X$ and $p$ is the probability density function, then \begin{align} L_{F_{X}}(s) = \frac{L_{p_{X}}(s)}{s} \end{align} which can be proven using integration by parts as follows \begin{align} L_{F_{X}}(s) = \int_{\mathbb{R}^{\geq 0}} F_{X}(x) \frac{d(e(-sx))}{s} = 0 + \frac{1}{s} \int_{\mathbb{R}^{\geq 0}} p_{X}(x) e^{-sx} dx \end{align}
Armed with the concept of a Laplace transform, we make the following observation first \begin{align} u(t) &= \sum_{i=1}^{\infty} F^{n*}(t) = F(t) + \sum_{i=2}^{\infty} F^{n*}(t)\newline &= F(t) + \big( \sum_{i=1}^{\infty} F^{n*}(t) \big) * F(t)\newline &= F(t) + u(t) * F(t)\newline u(t) &= F(t) + u(t) * p(t) \end{align} where $p$ is the probability density function and the last line stems from the fact that $\int u * F = \int u(x-y) dF(y) = \int u(x-y) p(y) dy$. Taking Laplace transform on both sides,
\begin{align} L_{u}(s) &= L_{F}(s) + L_{u}(s) L_{p}(s)\newline L_{u}(s) &= \frac{L_{p}(s)}{s} + L_{u}(s) L_{p}(s) \text{ from point 3 above}\newline L_{u(s)} &= \frac{L_{p}(s)}{s(1-L_{p}(s))} \end{align}
The last equation can be used to calculate the laplace transform of $u(t)$ and consecutively guess the functional form of $u(t)$.
The following two theorems hold true for Renewal processes
If $E[\xi] = \mu < \infty$, then \begin{align} \lim_{t \to \infty} \frac{N_{t}}{t} = \frac{1}{\mu} \end{align} which is analogous to the strong law of large numbers. This can be proven as follows \begin{align} S_{N_{t}} \leq t \leq S_{N_{t} + 1} \text{ from the definition of $N_{t}$}\newline \text{or, } \frac{N_{t}}{S_{N_{t} + 1}} \leq \frac{N_{t}}{t} \leq \frac{N_{t}}{S_{N_{t}}} \end{align} we can calculate the limits on the two bounds as \begin{align} \lim_{t \to \infty} \frac{N_{t}}{S_{N_{t}}} = \lim_{n \to \infty} \frac{n}{S_{n}} = \frac{1}{\mu} \end{align} from the strong law of large numbers applied to $\lim_{n \to \infty} \frac{S_{n}}{n}$. Similarly, one can show \begin{align} \lim_{t \to \infty} \frac{N_{t}}{S_{N_{t} + 1}} = \lim_{t \to \infty} \frac{N_{t}}{N_{t} + 1} \lim_{t \to \infty} \frac{N_{t} + 1}{S_{N_{t} + 1}} = 1 * \frac{1}{\mu} \end{align}
If $Var(\xi) = \sigma^{2} < \infty$, then \begin{align} \lim_{t \to \infty} \frac{N_{t} - t/\mu}{\sigma \sqrt{t}/\mu^{3/2}} = \mathcal{N}(0,1) \end{align} which is analogous to the central limit theorem. It can be proven by considering the CLT on $\xi$s \begin{align} \lim_{n \to \infty} P(\frac{S_{n} - n\mu}{\sigma \sqrt{n}} \leq x) &= \text{CDF of }\mathcal{N}(0,1)\newline \text{or, } \lim_{n \to \infty} P(S_{n} \leq n\mu + \sigma \sqrt{n} x) &= \text{CDF of }\mathcal{N}(0,1)\newline \text{or, } \lim_{n \to \infty} P(N_{t} \geq n) &= \text{CDF of }\mathcal{N}(0,1) \text{ from definition of $N_{t}$, where $t = n\mu + \sigma \sqrt{n} x$}\newline \end{align} We substitute $n\mu = t$ for very large value of $n$, since the total time will become total variables into the expected time for one $\xi$ when $n$ is large. Hence, \begin{align} n &= \frac{t}{\mu} - \frac{\sigma \sqrt{t}}{\mu^{3/2}}x\newline \lim_{n \to \infty} P(N_{t} \geq n) &= \lim_{n \to \infty} P(\frac{N_{t} - t/\mu}{\sigma \sqrt{t}/\mu^{3/2}} \leq x) = \text{CDF of }\mathcal{N}(0,1) \end{align}