Answer
Converges absolutely
Work Step by Step
Let us suppose a series $a_k$ such that $L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|$ converges when $L\lt 1$ and diverges when $L \gt 1$.
Now, we have: $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty} \dfrac{(k+1)^3}{e^{k+1}} \times \dfrac{e^k}{k^3}$
Use L-hospitals Rule: $L=\lim\limits_{k \to \infty} |\dfrac{-6}{6e}|\\=\lim\limits_{k \to \infty} |\dfrac{-1}{e}| \\=\dfrac{1}{e} \lt 1$
So, we can conclude that the given series converges absolutely.
