Group and a sequence of order of elements 
Give an example of a group $G$ such that the sequence $s_n= \dfrac{o(g_1) +o(g_2)+...+o(g_n)}{n}$ ($o(g_n)$ is the order of element $g_n$) has $\lim_{n\to\infty} s_n \in \mathbb Q^c$ ($\mathbb Q^c $ denotes the irrational numbers).

Now we know in math analysis that if ${a_n \to a}$ then $s_n \to a$ (note that  $s_n =\frac {a_1 +a_2 +...+a_n} {n}$). if $s_n \to s $ it don't imply $a_n \to s$ ( consider $a_n =(-1)^n$ and $a_n$ is divergent but $s_n \to 0$). we know that order of element of group is a nature number and $\Bbb{N}$ with absolute meter is equal discrete topology. we know in discrete topology every sequence is convergence iff it be constant. I consider for this question group $\frac{\Bbb{Q} }{\Bbb{Z}}$ because this group is infinite and for every nature N has element with order n. but I can't give a sequence in this group with above properties.
 A: Let $S\neq \emptyset$ and let $(G_s)_{s\in S}$ be a family of groups. The product group $$P=\prod_{s\in S}G_s$$ is the set of functions $f$ on $S$ with $f(s)\in G_s$ for each $s\in S$. The group operation on $P$ is $$(fg)(s)=f(s)g(s)$$ for each $s\in S$, when $f,g\in S$, where $f(s)g(s)$ denotes the product in the group $G_s$. Now let $S$ be the positive integers and for $n\in S$ let $G_n$ be a cyclic group with $x_n$ elements with $x_n\geq 2$, and let $y_n$ be a generator for $G_n$, so the order of $y_n$ in $G_n$ is $x_n$. For each $n$ let $1_n$ be the identity of $G_n$. For each $n$ we will let $$g_n\in P=\prod_{n\in S}G_s$$ where $$(m\ne n \implies g_n(m)=1_m)\wedge (g_n(n)=y_n).$$ In $P$ we have $\mathrm{ord}(g_n)=x_n$. 
Now we show that we can select a sequence $(x_n)$ so that $r=\lim_{n\to \infty}n^{-1}\sum_{j=1}^{n}x_j$ exists and can be any value $r\geq 2$. Let $[r]$ be the largest integer not exceeding $r$. Let $x_1=[r]$. Inductively, let $$x_{n+1}=[r] \text { if } n^{-1}\sum_{j=1}^{n}x_j\geq r ,$$ $$\text {and } x_{n+1}=[r]+1\text { if } n^{-1}\sum_{j=1}^{n}x_j<r.$$ I leave it as an exercise that $\lim_{n\to \infty}n^{-1}\sum_{j=1}^{n}x_j=r.$
