Let $p_j$ be the prime numbers in order, indexed by $j$. Define $S_n$ as follows:
\begin{equation} \begin{aligned} S_1 &:= \mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / 2 \mathbb Z &\oplus &\cdots \\ S_2 &:= \mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\cdots \\ S_3 &:= \mathbb Z / 2 \mathbb Z &\oplus &\mathbb Z / (2\cdot 3 \mathbb Z) &\oplus &\mathbb Z / (2 \cdot 3 \cdot 5 \mathbb Z) &\oplus &\mathbb Z / (2 \cdot 3 \cdot 5 \mathbb Z) &\oplus &\cdots \\ \cdots \end{aligned} \end{equation} We have an obvious inverse system $\{S_n, f_n\}$, where $S_n \xrightarrow {f_n} S_{n-1}$.
What, then, is the inverse limit $\varprojlim S_n =: S$?
I am not sure how to describe it. Clearly, it is not $P := \prod_{0<i}^\infty \left( \mathbb Z / \left( \left(\prod_{0<j}^i p_j\right)\mathbb Z\right)\right)$, as $(1,1,1,\cdots)$ is an element of P, but not of $S$ (note that $S_n$ is an infinite direct sum, not an infinite direct product). Also, it is not $S' := \bigoplus_{0<i}^\infty \left( \mathbb Z / \left( \left(\prod_{0<j}^i p_j\right)\mathbb Z\right)\right)$, as $(1,2,6,30,210,...,\prod_{0<j}^n p_j,...)$ does not belong to $S'$, but it does belong to $S$, since the element's restriction to any $S_n$ is a finite sum.
So apparently it is something in-between, and I wonder whether anybody knows of a nice description.
So far I have failed to do any better than something in English, such as: "The elements of $P$ of which, for any prime $p$, only finitely many coordinates are not divisible by $p$".
Using the Chinese remainder theorem to split everything as a direct sum of cyclic groups of prime order, you can describe each $S_n$ as just a direct sum of countably infinitely many copies of $\mathbb{Z}/p\mathbb{Z}$ for each of the first $n$ primes $p$, with $f_n$ just being the projection that drops the last prime. This gives a simple description of $S$ as $$\prod_{p\text{ prime}}(\mathbb{Z}/p\mathbb{Z})^{\oplus\mathbb{N}}.$$