Math Problem Statement

Can you tell me it's interval of convergence?

Solution

To determine the interval of convergence of the power series:

S(x)=n=0(n+1)(n+2)n!xn,S(x) = \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{n!} x^n,

we can use the Ratio Test. The Ratio Test involves finding the limit of the absolute value of the ratio of successive terms of the series. For a power series anxn\sum a_n x^n, the series converges if:

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

Here, the nn-th term ana_n is given by:

an=(n+1)(n+2)n!.a_n = \frac{(n+1)(n+2)}{n!}.

The (n+1)(n+1)-th term an+1a_{n+1} is:

an+1=(n+2)(n+3)(n+1)!.a_{n+1} = \frac{(n+2)(n+3)}{(n+1)!}.

The ratio of successive terms is:

an+1an=(n+2)(n+3)(n+1)!(n+1)(n+2)n!\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\frac{(n+2)(n+3)}{(n+1)!}}{\frac{(n+1)(n+2)}{n!}}\right|

=(n+2)(n+3)(n+1)!n!(n+1)(n+2)= \left|\frac{(n+2)(n+3)}{(n+1)!} \cdot \frac{n!}{(n+1)(n+2)}\right|

=(n+3)(n+1).= \left|\frac{(n+3)}{(n+1)}\right|.

As nn approaches infinity, the ratio simplifies to:

limnn+3n+1=limnn(1+3n)n(1+1n)=1.\lim_{n \to \infty} \left|\frac{n+3}{n+1}\right| = \lim_{n \to \infty} \left|\frac{n(1 + \frac{3}{n})}{n(1 + \frac{1}{n})}\right| = 1.

Since the limit equals 1, the Ratio Test is inconclusive at this stage. This implies that the series might converge only at certain points, particularly where xx affects the convergence.

Let's reapply the ratio with the xnx^n term included:

limnan+1xn+1anxn=limnn+3n+1x\lim_{n \to \infty} \left|\frac{a_{n+1}x^{n+1}}{a_n x^n}\right| = \lim_{n \to \infty} \left|\frac{n+3}{n+1}\right| |x|

=xlimnn+3n+1=x.= |x| \lim_{n \to \infty} \left|\frac{n+3}{n+1}\right| = |x|.

The series converges if x<1|x| < 1 and diverges if x>1|x| > 1. The case x=1|x| = 1 needs separate checking by other convergence tests (such as the Root or Ratio Test applied carefully or by direct evaluation of the series), but typically for series involving factorials, the boundary points x=1|x| = 1 do not contribute to convergence.

Thus, the interval of convergence is:

x<1,or(1,1).|x| < 1, \quad \text{or} \quad (-1, 1).

For the full determination, further tests at x=1x = 1 and x=1x = -1 are needed, but generally, these points are not included without further evaluation.

Would you like a deeper dive into why these boundary points might be excluded or included, or any other clarification? Here are some questions that could be helpful:

  1. How does the Ratio Test determine convergence?
  2. What are some exceptions where the Ratio Test might not give conclusive results?
  3. What is the significance of the interval of convergence for power series?
  4. How do we handle boundary points in the interval of convergence?
  5. How can we use other convergence tests alongside the Ratio Test?

Tip: When using the Ratio Test, always consider evaluating the series at the endpoints separately, as the test does not provide information about convergence at the boundary.

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Math Problem Analysis

Mathematical Concepts

Power series
Series convergence
Ratio Test

Formulas

Ratio Test for series convergence

Theorems

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Suitable Grade Level

Advanced undergraduate