The problem asks us to analyze the sequence $a_n = e^{-n} \sin n$. We need to find the first few terms, determine if the sequence converges, and if it converges, find the limit as $n$ approaches infinity.

1. Problem Description

The problem asks us to analyze the sequence $a_n = e^{-n} \sin n$. We need to find the first few terms, determine if the sequence converges, and if it converges, find the limit as $n$ approaches infinity.

2. Solution Steps

First, let's find the first few terms of the sequence.

$a_1 = e^{-1} \sin 1 \approx 0.3096$

$a_2 = e^{-2} \sin 2 \approx 0.1231$

$a_3 = e^{-3} \sin 3 \approx 0.0212$

$a_4 = e^{-4} \sin 4 \approx -0.0140$

$a_5 = e^{-5} \sin 5 \approx -0.0192$

Now, let's determine if the sequence converges. We have $a_n = e^{-n} \sin n$. We can rewrite this as $a_n = \frac{\sin n}{e^n}$.

Since $-1 \le \sin n \le 1$, we have $-\frac{1}{e^n} \le \frac{\sin n}{e^n} \le \frac{1}{e^n}$.

As $n \to \infty$, $e^n \to \infty$, so $\frac{1}{e^n} \to 0$.

Thus, $-\frac{1}{e^n} \to 0$ and $\frac{1}{e^n} \to 0$.

By the Squeeze Theorem, $\lim_{n \to \infty} \frac{\sin n}{e^n} = 0$.

3. Final Answer

The first few terms are approximately $0.3096, 0.1231, 0.0212, -0.0140, -0.0192$.

The sequence converges to

0. $\lim_{n \to \infty} a_n = 0$