Answer
converges to $e^{2/3}$
Work Step by Step
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} (1+\dfrac{x}{n})^{n}=e^x$
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{3n+1}{3n-1})^{n}$
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (1+\dfrac{2}{3n-1})^{n}=e^2$
Let us consider $f(x)=x^{(\frac{n}{3n-1})} $ so, $ f(e^2)=(e^2)^{({\frac{n}{3n-1}})}$
or, $\lim\limits_{n \to \infty} (e^2)^{({\frac{n}{3n-1}})}=e^{2/3}$
Thus, $\lim\limits_{n \to \infty} a_n=e^{2/3}$ and {$a_n$} converges to $e^{2/3}$.
