Suppose $G$ is a group. $g \in G$. Define $\{C_G^{(n)}(g)\}_{n = 0}^\infty$ as a series of subgroups of $G$, defined by the following recurrence relation:
$$C_G^n(g) = \begin{cases} \langle \{h \in G| [h, g] \in C_G^{n-1}\} \rangle & \quad n \geq 1 \\ E & \quad n = 0 \end{cases}$$
It is not hard to see, that this series is ascending and that $C_G^1(g) = C_G(g)$ is the centraliser of $g$ in $G$.
The question is:
Is it always true, that $C_G^n(g) \triangleleft C_G^{n + 1}(g)$?
For $n = 0$ it is obviously true. For $n > 0$ it is sufficient to prove, that $\forall a \in C_G^n(G) \forall b \in G ([b, g] \in C_G^n(g) \to b^{-1}ab \in C_G^n(g))$. However, this does not seem to be any easier.
No. This is false.
If it were true, then it would have resulted in all left-Engel elements of the group to lie in subnormal abelian subgroups and thus in the locally-nilpotent radical of the corresponding group. However not all left-Engel elements satisfy that condition, as proven in “Left $3$-Engel elements in groups of exponent $5$” by G. Traustason.