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To test the convergence of the series

\[
\sum_{n=1}^{\infty} \frac{\sin\left(\frac{n \pi}{2}\right)}{n!}
\]

using the Ratio Test, we need to evaluate the limit:

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

where \( a_n = \frac{\sin\left(\frac{n \pi}{2}\right)}{n!} \).

### Step 1: Set up \( a_n \) and \( a_{n+1} \)

1. The \( n \)-th term of the series is:
   \[
   a_n = \frac{\sin\left(\frac{n \pi}{2}\right)}{n!}.
   \]

2. The \((n+1)\)-th term is:
   \[
   a_{n+1} = \frac{\sin\left(\frac{(n+1) \pi}{2}\right)}{(n+1)!}.
   \]

### Step 2: Formulate \( \frac{a_{n+1}}{a_n} \)

We find:
\[
\frac{a_{n+1}}{a_n} = \frac{\sin\left(\frac{(n+1) \pi}{2}\right)}{(n+1)!} \cdot \frac{n!}{\sin\left(\frac{n \pi}{2}\right)} = \frac{\sin\left(\frac{(n+1) \pi}{2}\right)}{\sin\left(\frac{n \pi}{2}\right)} \cdot \frac{1}{n+1}.
\]

### Step 3: Analyze the Behavior of \( \sin\left(\frac{n \pi}{2}\right) \)

Notice that \(\sin\left(\frac{n \pi}{2}\right)\) alternates in a cycle with \( n \) as follows:

- For \( n = 1 \): \(\sin\left(\frac{\pi}{2}\right) = 1\)
- For \( n = 2 \): \(\sin(\pi) = 0\)
- For \( n = 3 \): \(\sin\left(\frac{3\pi}{2}\right) = -1\)
- For \( n = 4 \): \(\sin(2\pi) = 0\)

Thus, the terms with \( n = 2, 4, 6, \ldots \) (even \( n \)) are zero, meaning these terms contribute nothing to the series.

### Step 4: Apply the Ratio Test for Non-Zero Terms

For the non-zero terms (odd \( n \)), we have:
\[
\left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{n+1} \quad \text{(for odd \( n \))}.
\]

Taking the limit as \( n \to \infty \):
\[
L = \lim_{n \to \infty} \frac{1}{n+1} = 0.
\]

### Conclusion

Since \( L = 0 < 1 \), the Ratio Test confirms that the series **converges**.

---

Would you like a deeper explanation or have any questions on this solution?

### Related Questions

1. What is the Ratio Test, and why does it work?
2. Can the Alternating Series Test also be used here?
3. How does factorial growth influence the convergence of series?
4. What is the behavior of trigonometric functions in series like these?
5. Can you apply the Ratio Test to any type of series?

### Tip

When applying the Ratio Test, if the limit approaches zero, the series is guaranteed to converge, regardless of other conditions.

Use the ratio test to test for convergence of the series ∑ (sin(nπ/2)) / n! from n=1 to ∞.

Analyze the convergence of the series ∑ (sin(nπ/2)) / n! using the Ratio Test. Understand the impact of trigonometric functions and factorial growth in determining convergence. Suitable for advanced college-level calculus.

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