The problem asks us to analyze the sequence $a_n = e^{-n} \sin n$. We need to find the first few terms of the sequence, determine if it converges or diverges, 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 of the sequence, determine if it converges or diverges, 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:

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

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

$a_3 = e^{-3} \sin 3 \approx 0.0142$

$a_4 = e^{-4} \sin 4 \approx -0.0149$

$a_5 = e^{-5} \sin 5 \approx -0.0128$

$a_6 = e^{-6} \sin 6 \approx -0.0009$

Now, let's analyze the convergence of the sequence. We have $a_n = e^{-n} \sin n$. We can rewrite this as $a_n = \frac{\sin n}{e^n}$. We know that $-1 \le \sin n \le 1$ for all $n$. Therefore, we have:

$-\frac{1}{e^n} \le \frac{\sin n}{e^n} \le \frac{1}{e^n}$.

As $n$ approaches infinity, $e^n$ also approaches infinity. Therefore, $\lim_{n \to \infty} \frac{1}{e^n} = 0$ and $\lim_{n \to \infty} -\frac{1}{e^n} = 0$.

By the squeeze theorem, since $-\frac{1}{e^n} \le \frac{\sin n}{e^n} \le \frac{1}{e^n}$ and $\lim_{n \to \infty} -\frac{1}{e^n} = 0$ and $\lim_{n \to \infty} \frac{1}{e^n} = 0$, we have:

$\lim_{n \to \infty} \frac{\sin n}{e^n} = 0$.

Therefore, the sequence $a_n = e^{-n} \sin n$ converges to

0.

3. Final Answer

The first few terms are approximately: 0.3096, 0.1231, 0.0142, -0.0149, -0.0128, -0.

0

0

0

9. The sequence converges.

The limit of the sequence is

0.