## University Calculus: Early Transcendentals (3rd Edition)

Converges to $3$.
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 $3$.