Is it possible for a finite dimensional $k$-algebra $A$ ($k$ an algebraically closed field of characteristic $p$) to have an infinite socle series?
Namely, I have calculated the radical and socle for $T_n(k)=$ the lower triangular matrices. I found (I will demonstrate the $3 \times 3$ case
$$rad T_n(k) = \begin{pmatrix} 0 & 0 & 0\\ x & 0 & 0\\ y & z &0 \end{pmatrix}$$
and so
$$rad^2 T_n(k) = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ y & 0 &0 \end{pmatrix}$$
and $rad^3 T_n(k) = 0$. If we consider the simple modules or, equivalently, consider what is annihiliated by radical of $T_n(k)$ we come up with the diagonal matrices
$$soc T_n(k) = \begin{pmatrix} a & 0 & 0\\ 0 & b & 0\\ 0 & 0 & c \end{pmatrix}$$
which is clearly not nilpotent. I was under the impression the socle series was supposed to terminate but it seems this is a solid counter example unless I have made a mistake.
The socle series stabilizes in any Artinian ring, just like any ascending chain.
A finite dimensional $k$-algebra is (two-sided) Artinian, because every left or right ideal is a $k$-vector subpace.
The socle of a ring is a two-sided ideal, so what you found is not the socle.