Answer
converges to $0$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n=\dfrac{\ln (2n^3+1)}{n}$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}[\dfrac{\ln (2n^3+1)}{n}]$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{\infty}{\infty}$
Since, we can see that the limit has the form of $\frac{\infty}{\infty}$, use L-Hospital's rule.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{\frac{6n^2}{(2n^3+1)}}{1}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{6n^2}{(2n^3+1)}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{6}{(2n^3/n^2+1/n^2)}=\frac{6}{\infty}$
$\lim\limits_{n \to \infty}a_n=0$
Therefore, the sequence converges to $0$.
