Math Problem Statement

Interval of convergence for the sum k=0 to Infinity of ((-1)^k x^k)/(2k)!)

Solution

To determine the interval of convergence for the series:

k=0(1)kxk(2k)!\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{(2k)!}

We can use the Ratio Test for convergence. The Ratio Test states that for a series ak\sum a_k, the series converges absolutely if:

L=limkak+1ak<1L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1

If L>1L > 1, the series diverges. If L=1L = 1, the test is inconclusive, and we need to investigate further.

Step-by-Step Solution

  1. General Term: The general term in the series is:

    ak=(1)kxk(2k)!a_k = \frac{(-1)^k x^k}{(2k)!}

  2. Applying the Ratio Test: We calculate the limit of the ratio of successive terms:

    ak+1ak=(1)k+1xk+1(2(k+1))!(2k)!(1)kxk\frac{a_{k+1}}{a_k} = \frac{(-1)^{k+1} x^{k+1}}{(2(k+1))!} \cdot \frac{(2k)!}{(-1)^k x^k}

    Simplifying this expression:

    ak+1ak=(1)k+1(1)kxk+1xk(2k)!(2k+2)!\frac{a_{k+1}}{a_k} = \frac{(-1)^{k+1}}{(-1)^k} \cdot \frac{x^{k+1}}{x^k} \cdot \frac{(2k)!}{(2k+2)!}

    ak+1ak=(1)x1(2k+1)(2k+2)\frac{a_{k+1}}{a_k} = (-1) \cdot x \cdot \frac{1}{(2k+1)(2k+2)}

    Thus,

    ak+1ak=x(2k+1)(2k+2)\left| \frac{a_{k+1}}{a_k} \right| = \frac{|x|}{(2k+1)(2k+2)}

  3. Taking the Limit: As kk \to \infty, the denominator (2k+1)(2k+2)(2k+1)(2k+2) grows without bound, so:

    limkx(2k+1)(2k+2)=0\lim_{k \to \infty} \frac{|x|}{(2k+1)(2k+2)} = 0

  4. Conclusion from Ratio Test: Since the limit is 00 for all values of xx, the Ratio Test tells us that the series converges for all xx.

Interval of Convergence:

The series converges for all real values of xx, so the interval of convergence is:

(,)(-\infty, \infty)

Additional Notes:

  • The series converges absolutely for all xx, as shown by the Ratio Test.
  • This result is similar to the series for the exponential function, though this series involves only even powers of xx.

Would you like more details on this approach? Or do you have other questions?

Here are some related questions to explore further:

  1. How does the Ratio Test apply to other series like xnn!\sum \frac{x^n}{n!}?
  2. What are the conditions for applying the Root Test for series convergence?
  3. How can you determine the radius of convergence for a power series?
  4. What are the implications of convergence at the boundary points for series?
  5. Can we apply the alternating series test to determine convergence of this series?

Tip: When using the Ratio Test, if the limit approaches zero, the series always converges for any value of xx.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Power Series
Ratio Test
Factorials

Formulas

a_k = (-1)^k x^k / (2k)!
L = lim(k->∞) |a_(k+1) / a_k|
|a_(k+1) / a_k| = |x| / ((2k+1)(2k+2))

Theorems

Ratio Test

Suitable Grade Level

Grades 11-12