Math Problem Statement

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

Solution

To test the convergence of the series

n=1sin(nπ2)n!\sum_{n=1}^{\infty} \frac{\sin\left(\frac{n \pi}{2}\right)}{n!}

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

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

where an=sin(nπ2)n!a_n = \frac{\sin\left(\frac{n \pi}{2}\right)}{n!}.

Step 1: Set up ana_n and an+1a_{n+1}

  1. The nn-th term of the series is: an=sin(nπ2)n!.a_n = \frac{\sin\left(\frac{n \pi}{2}\right)}{n!}.

  2. The (n+1)(n+1)-th term is: an+1=sin((n+1)π2)(n+1)!.a_{n+1} = \frac{\sin\left(\frac{(n+1) \pi}{2}\right)}{(n+1)!}.

Step 2: Formulate an+1an\frac{a_{n+1}}{a_n}

We find: an+1an=sin((n+1)π2)(n+1)!n!sin(nπ2)=sin((n+1)π2)sin(nπ2)1n+1.\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(nπ2)\sin\left(\frac{n \pi}{2}\right)

Notice that sin(nπ2)\sin\left(\frac{n \pi}{2}\right) alternates in a cycle with nn as follows:

  • For n=1n = 1: sin(π2)=1\sin\left(\frac{\pi}{2}\right) = 1
  • For n=2n = 2: sin(π)=0\sin(\pi) = 0
  • For n=3n = 3: sin(3π2)=1\sin\left(\frac{3\pi}{2}\right) = -1
  • For n=4n = 4: sin(2π)=0\sin(2\pi) = 0

Thus, the terms with n=2,4,6,n = 2, 4, 6, \ldots (even nn) 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 nn), we have: \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{n+1} \quad \text{(for odd n)}.

Taking the limit as nn \to \infty: L=limn1n+1=0.L = \lim_{n \to \infty} \frac{1}{n+1} = 0.

Conclusion

Since L=0<1L = 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.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Ratio Test
Factorials
Trigonometric Functions in Series

Formulas

Ratio Test formula L = lim(n→∞) |a_(n+1) / a_n|
Sine function sin(nπ/2) for alternating series

Theorems

Ratio Test for Series Convergence

Suitable Grade Level

College Level