The problem asks us to test the convergence of the series

$\sum_{n=1}^{\infty} \frac{\sin(n)}{n!}$

using the Ratio Test.

Solution Using the Ratio Test

The Ratio Test states that for a series $\sum a_n$, if

$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L,$

then:

• If $L < 1$, the series converges absolutely.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

In this problem, we have: $a_n = \frac{\sin(n)}{n!}.$

Let's apply the Ratio Test by finding $\left| \frac{a_{n+1}}{a_n} \right|$:

$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\sin(n+1)}{(n+1)!} \cdot \frac{n!}{\sin(n)} \right| = \left| \frac{\sin(n+1)}{\sin(n)} \cdot \frac{1}{n+1} \right|.$

Step-by-Step Analysis:

1. As $n \to \infty$, $\sin(n)$ oscillates between -1 and 1, so $\sin(n)$ and $\sin(n+1)$ do not grow without bound.
2. However, the factorial in the denominator, $(n+1)!$, grows very rapidly as $n$ increases.

Because the factorial growth in the denominator (from $\frac{1}{n+1}$) dominates the oscillations of $\sin(n)$, we expect this limit to approach zero as $n \to \infty$.

Thus: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1.$

Conclusion

Since $L = 0 < 1$, the Ratio Test confirms that the series $\sum_{n=1}^{\infty} \frac{\sin(n)}{n!}$ converges absolutely.

Would you like further details, or do you have any questions?

Additional Related Questions

1. What happens if we test the convergence of $\sum_{n=1}^{\infty} \frac{\cos(n)}{n!}$ instead?
2. Can the Ratio Test be used on series with only trigonometric terms (e.g., $\sum_{n=1}^{\infty} \sin(n)$)?
3. How does the factorial in the denominator influence convergence in general?
4. What is an example of a divergent series that fails the Ratio Test?
5. How would we test convergence if we had $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2}$ instead?

Tip

The Ratio Test is especially useful when dealing with factorials or exponential terms in a series, as these terms tend to grow (or decay) very quickly, making it easier to apply the test effectively.