Let $p$ be a prime in $\mathbb{Z}$. The p-adic order or p-adic valuation for $\mathbb{Z}$ is defined as $\nu_p : \mathbb{Z} \rightarrow \mathbb{N}$ such that \begin{equation*} \nu_p(n) =\begin{cases} \max\{v \in \mathbb{N} : p^{v}|n \} & \text{if } n \neq 0\\ \infty & \text{if } n=0\\ \end{cases} \end{equation*}
Now, consider $S$ to be a set of uniformly distributed random integers. Then we have $$ \nu_p(S) = \frac{1}{p} + \frac{1}{p^2} + \dots = \frac{1}{p-1}$$
I have been trying to figure this out for a while, but I don't how ? Am I missing something obvious here? Any hints would be welcome !
It appears to me that $\nu_p(S)$ is intended to be the expected $p$-adic valuation of a member of $S$ chosen at random with uniform probability. I’m no probabilist, but the reasoning must be something along the following lines.
Say $S$ is finite, of cardinality $n$. For $k\ge 1$ you expect about $\frac{n}{p^k}$ members of $S$ to be divisible by $p^k$. If $s\in S$ and $\nu_p(s)=m$, $s$ is divisible by $p^k$ for $k=1,\ldots,m$, so
$$\begin{align*} \sum_{s\in S}\nu_p(s)&=\sum_{s\in S}\sum_{k=1}^{\nu_p(s)}1\\ &=\sum_{k\ge 1}\sum_{s\in S}\left[p^k\mid s\right]\\ &\approx\sum_{k\ge 1}\frac{n}{p^k}\;, \end{align*}$$
and the expected $p$-adic valuation of a randomly chosen member of $S$ is therefore approximately
$$\frac1n\sum_{k\ge 1}\frac{n}{p^k}=\sum_{k\ge 1}\frac1{p^k}=\frac1{p-1}\;.$$
(Here $\left[p^k\mid s\right]$ is an Iverson bracket.)