Math Problem Statement

Determine the interval of convergence of the function and analyze the behavior of the series at the endpoints of the interval

Determine the interval of convergence of the function and analyze the behavior of the series at the endpoints of the interval.

Solution

The given series is:

n=1n!nn(x+3)n\sum_{n=1}^{\infty} \frac{n!}{n^n}(x+3)^n

Step 1: Applying the Ratio Test for Convergence

To determine the interval of convergence, we can apply the Ratio Test. This test is suitable for power series and requires evaluating:

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

where an=n!nn(x+3)na_n = \frac{n!}{n^n}(x+3)^n. Now, compute the ratio of consecutive terms:

an+1an=(n+1)!(n+1)n+1(x+3)n+1n!nn(x+3)n\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}(x+3)^{n+1}}{\frac{n!}{n^n}(x+3)^n}

This simplifies to:

an+1an=(n+1)!(n+1)n+1nnn!(x+3)\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \cdot (x+3)

The factorial terms cancel out, and we are left with:

=(n+1)(n+1)n+1nn(x+3)= \frac{(n+1)}{(n+1)^{n+1}} \cdot n^n \cdot (x+3)

This can be further simplified as:

=nn(n+1)n1n+1(x+3)= \frac{n^n}{(n+1)^n} \cdot \frac{1}{n+1} \cdot (x+3)

=(nn+1)n1n+1(x+3)= \left( \frac{n}{n+1} \right)^n \cdot \frac{1}{n+1} \cdot (x+3)

Step 2: Taking the Limit as nn \to \infty

As nn \to \infty, (nn+1)n\left( \frac{n}{n+1} \right)^n approaches 1e\frac{1}{e}, and 1n+10\frac{1}{n+1} \to 0. Therefore, the entire expression for the ratio becomes:

L=limn1e(x+3)0=0L = \lim_{n \to \infty} \left| \frac{1}{e} \cdot (x+3) \cdot 0 \right| = 0

This shows that the series converges for all real values of xx. Therefore, the interval of convergence is (,)(-\infty, \infty).

Step 3: Behavior at the Endpoints

Since the ratio test shows that the series converges for all xx, there is no need to check the behavior at specific endpoints, as there are no boundaries for this case.

Final Conclusion:

The interval of convergence of the given series is (,)(- \infty, \infty), and the series converges everywhere on the real line.

Would you like more details on this or have any other questions?

5 Related Questions:

  1. How would the interval of convergence change if the factorial term was different?
  2. Can the root test be applied to the same series, and how would the results compare?
  3. What happens if the term nnn^n is replaced with nkn^k, where kk is a constant?
  4. How does the ratio test compare with other tests for determining convergence?
  5. What is the significance of the factor (x+3)n(x+3)^n in determining convergence?

Tip:

Always simplify the ratio as much as possible before applying limits to ensure accuracy when using the ratio test for power series.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Interval of Convergence
Power Series

Formulas

Ratio Test: L = lim(n -> ∞) |a(n+1)/a(n)|
Given series: ∑ (n! / n^n) (x+3)^n

Theorems

Ratio Test
Power Series Convergence

Suitable Grade Level

Undergraduate Mathematics