## Calculus (3rd Edition)

The sequence $y_n$ diverges.
We have $$\lim_{n\to \infty}y_n=\lim_{n\to \infty} \frac{e^n}{2^n} \\ = \lim_{n\to \infty} (e/2)^n = \infty .$$ Since we used the fact that $e/2\gt 1.$ Hence, by Theorem 1, the sequence $y_n$ diverges.