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

Converges to $5$
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (3^n +5^n)^{1/n}$ Since, $\lim\limits_{n \to \infty} x^{1/n}=1$ when $x \gt 0$ So, $\lim\limits_{n \to \infty} 5((\dfrac{3}{5})^n+1)^n=5$ Hence, $\lim\limits_{n \to \infty} a_n=5$ and {$a_n$} is Convergent and converges to $5$