Answer
converges to $3$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n=\sqrt[n] \frac{3^n}{n}$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\sqrt[n] \frac{3^n}{n}$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\sqrt[n] \frac{3^n}{n}=\lim\limits_{n \to \infty}(\dfrac{3^n}{n})^{1/n}$
$\lim\limits_{n \to \infty}a_n=\dfrac{\lim\limits_{n \to \infty}3}{\lim\limits_{n \to \infty}n^{1/n}}$ ...(1)
Let us consider the denominator term such as:
$y=\lim\limits_{n \to \infty}n^{1/n}$
Use logarithmic rule: $\ln a^n=n \ln a$
$\ln y=\lim\limits_{n \to \infty} \frac{\ln n}{n}$
Since, we can see that the limit has the form of $\frac{\infty}{\infty}$, so take the help of L-Hospital's rule.
$\ln y=\lim\limits_{n \to \infty} \frac{1/ n}{n}$
$\ln y=0$
$e^{\ln y}=e^0 \implies y=1$
From equation (1), we have $\lim\limits_{n \to \infty}a_n=3$
Therefore, the sequence converges to $3$.
