Answer
Absolutely convergent.
Work Step by Step
A $p$-series has the form of $\Sigma_{n=k}^{\infty}\dfrac{1}{n^p}$. It is convergent iff $p \gt 1$ and otherwise diverges.
From the given problem, we have $\Sigma_{n=1}^{\infty}\dfrac{1}{2n^3}$
The given series can be re-written as: $a_n=\frac{1}{2}\Sigma_{n=1}^{\infty}\dfrac{1}{ n^{3}}$
Here, $p=3 \gt 1$
Thus, the given series is convergent. So, it is absolutely convergent.
