Answer
Diverges
Work Step by Step
Here, we are given that $a_1=\dfrac{1}{2}; a_{n+1}=\dfrac{4n-2}{3n+2} a_n$
Thus, $\dfrac{a_{n+1}}{a_n}=\dfrac{4n-2}{3n+2}$
Next, we have: $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} \dfrac{4n-2}{3n+2}=\dfrac{4}{3} \gt 1$
Hence, we can conclude that the given series diverges.
