Math Problem Statement

Test for convergence of the following series using the ratio test: sum of sin(πn/2) / n! from n = 1 to infinity.

Solution

The problem requires using the ratio test to check the convergence of the series:

n=1sin(π2n)n!\sum_{n=1}^{\infty} \frac{\sin\left(\frac{\pi}{2}n\right)}{n!}

Solution

To apply the ratio test, we consider the terms of the series:

an=sin(π2n)n!a_n = \frac{\sin\left(\frac{\pi}{2}n\right)}{n!}

The ratio test states that a series an\sum a_n converges absolutely if

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

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

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

  2. Find an+1an\frac{a_{n+1}}{a_n}: an+1an=sin(π2(n+1))(n+1)!n!sin(π2n)=sin(π2(n+1))sin(π2n)1n+1\frac{a_{n+1}}{a_n} = \frac{\sin\left(\frac{\pi}{2}(n+1)\right)}{(n+1)!} \cdot \frac{n!}{\sin\left(\frac{\pi}{2}n\right)} = \frac{\sin\left(\frac{\pi}{2}(n+1)\right)}{\sin\left(\frac{\pi}{2}n\right)} \cdot \frac{1}{n+1}

Step 2: Analyze sin(π2n)\sin\left(\frac{\pi}{2}n\right)

Notice that sin(π2n)\sin\left(\frac{\pi}{2}n\right) oscillates depending on the value of nn:

  • When n=1,5,9,n = 1, 5, 9, \dots, sin(π2n)=1\sin\left(\frac{\pi}{2}n\right) = 1.
  • When n=3,7,11,n = 3, 7, 11, \dots, sin(π2n)=1\sin\left(\frac{\pi}{2}n\right) = -1.
  • For even values of nn, sin(π2n)=0\sin\left(\frac{\pi}{2}n\right) = 0, which implies that some terms of the series are zero.

Step 3: Take the Limit as nn \to \infty

Since sin(π2(n+1))/sin(π2n)\sin\left(\frac{\pi}{2}(n+1)\right)/\sin\left(\frac{\pi}{2}n\right) oscillates and does not affect the factorial in the denominator, we focus on 1n+1\frac{1}{n+1}.

Thus, we get: limnan+1an=limn1n+1=0\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{1}{n+1} = 0

Conclusion

Since the limit is 0<10 < 1, the series converges absolutely by the ratio test.


Would you like further details or have any questions?

Related Questions

  1. How would the convergence be affected if the factorial term were removed from the denominator?
  2. Can the ratio test be used for all types of series?
  3. What other convergence tests could be applied to this series?
  4. How does the oscillating behavior of sin(π2n)\sin\left(\frac{\pi}{2}n\right) impact the series' convergence?
  5. What would happen if we replaced sin(π2n)\sin\left(\frac{\pi}{2}n\right) with another trigonometric function?

Tip: The ratio test is especially useful when terms involve factorials or exponential functions, as it simplifies the comparison of successive terms.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Ratio Test
Trigonometric Series
Factorials

Formulas

Ratio Test for series convergence
Factorial n!

Theorems

Ratio Test

Suitable Grade Level

College/University Level (Calculus)