I am studying residually nilpotent groups. I am using the following defintions.
Lower central series: Let $G$ be a group. Then we inductively define the lower central series of $G$ as $$\gamma_1(G) = G,\ \gamma_{i}(G) = [G,\gamma_{i-1}(G)]\ \forall i> 1.$$
Nilpotent: Let $G$ be a group. Then $G$ is said to be nilpotent if there exists an integer $c\geq 1$ such that $\gamma_c(G) = \{1\}$.
Residually nilpotent: Let $G$ be a group. Then we say that $G$ is residually nilpotent if $$ \bigcap_{n\geq 1} \gamma_n(G) = \{1\}.$$
It's clear that every nilpotent group is also residually nilpotent. However, the converse does not hold. For example, if $\mathcal{D}_\infty$ is the infinite dihedral group, then one can show that $\mathcal{D}_\infty$ is residually nilpotent but not nilpotent.
My question is the following: can we define similar concepts surrounding the upper central series of a group $G$. This series is defined as follows: $$ Z_0(G) = \{1\},\ \frac{Z_i(G)}{Z_{i-1}(G)} = Z\left(\frac{G}{Z_{i-1}(G)}\right)\ \forall i>0.$$
With the definition of residual nilpotency in mind, I was wondering if there exist non-nilpotent groups $G$ such that $$ \bigcup_{n\geq 0} Z_n(G) = G.$$ I can't seem to find any example of such a group, but I don't know why it wouldn't be possible. Has this counterpart of residually nilpotent groups been studied somewhere? I did some research, but could not find anything concrete.
Thanks in advance!
You could take an infinite direct product of finite nilpotent groups $G_n$ of class $n$, such as dihedral groups of order $2^{n+1}$. Note that any such example has to be infinitely generated.