Answer
Converges to $e^{2/3}$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{3n+1}{3n-1})^{n}$
Since, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$ when $x \gt 0$
So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{3n+1}{3n-1})^{n}= \lim\limits_{n \to \infty} (1+\dfrac{2}{3n-1})^{n}=e^2$
Suppose $f(x)=x^({\frac{n}{3n-1}}) \implies f(e^2)=(e^2)^{({\frac{n}{3n-1}})}$
or, $\lim\limits_{n \to \infty} (e^2)^{({\frac{n}{3n-1}})}=e^{2/3}$
Hence, $\lim\limits_{n \to \infty} a_n=e^{2/3}$ and {$a_n$} is convergent and converges to $e^{2/3}$
