Answer
The sequence $y_n$ diverges.
Work Step by Step
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.
