Math Problem Statement

Determine whether the series \( \sum_{n=1}^{\infty} \frac{n(x-4)^n}{n^3 + 1} \) is absolutely convergent or divergent using the ratio test.

Solution

To determine whether the series

n=1n(x4)nn3+1\sum_{n=1}^{\infty} \frac{n(x-4)^n}{n^3 + 1}

is absolutely convergent or divergent using the ratio test, we proceed as follows:

Step-by-Step Solution

  1. Identify the General Term: The general term of the series is an=n(x4)nn3+1.a_n = \frac{n(x-4)^n}{n^3 + 1}.

  2. Set Up the Ratio Test: The ratio test states that for a series an\sum a_n, we look at L=limnan+1an.L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.

    • If L<1L < 1, the series converges absolutely.
    • If L>1L > 1, the series diverges.
    • If L=1L = 1, the test is inconclusive.
  3. Calculate an+1an\left| \frac{a_{n+1}}{a_n} \right|: First, find an+1a_{n+1}: an+1=(n+1)(x4)n+1(n+1)3+1.a_{n+1} = \frac{(n+1)(x-4)^{n+1}}{(n+1)^3 + 1}.

    Now compute the ratio an+1an\frac{a_{n+1}}{a_n}: an+1an=(n+1)(x4)n+1(n+1)3+1n3+1n(x4)n.\frac{a_{n+1}}{a_n} = \frac{(n+1)(x-4)^{n+1}}{(n+1)^3 + 1} \cdot \frac{n^3 + 1}{n(x-4)^n}. Simplifying this expression: an+1an=(n+1)(x4)nn3+1(n+1)3+1.\frac{a_{n+1}}{a_n} = \left| \frac{(n+1)(x-4)}{n} \cdot \frac{n^3 + 1}{(n+1)^3 + 1} \right|.

  4. Take the Limit as nn \to \infty: Now, calculate the limit of this expression as nn \to \infty: L=limn(n+1)(x4)nn3+1(n+1)3+1.L = \lim_{n \to \infty} \left| \frac{(n+1)(x-4)}{n} \cdot \frac{n^3 + 1}{(n+1)^3 + 1} \right|.

    As nn \to \infty, n+1n1\frac{n+1}{n} \to 1 and n3+1(n+1)3+11\frac{n^3 + 1}{(n+1)^3 + 1} \to 1. So we get: L=x4.L = |x - 4|.

  5. Determine Convergence Based on LL:

    • For absolute convergence, we need L<1L < 1, which means x4<1|x - 4| < 1.
    • This simplifies to 3<x<53 < x < 5.

    Therefore, the series is absolutely convergent for 3<x<53 < x < 5.

Conclusion

The series is absolutely convergent for 3<x<53 < x < 5 and divergent otherwise.

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Further Related Questions

  1. What is the radius of convergence for this series?
  2. How would the convergence test change if we used the root test instead of the ratio test?
  3. For what values of xx does the series converge conditionally?
  4. How would we analyze the convergence if the series was centered around a different point?
  5. What is the effect of the nn-dependent numerator on the convergence?

Tip

When applying the ratio test, simplify each term as much as possible before taking the limit to avoid unnecessary complexity.

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Math Problem Analysis

Mathematical Concepts

Infinite Series
Ratio Test
Absolute Convergence

Formulas

Ratio test: \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)

Theorems

Ratio Test for Convergence of Series

Suitable Grade Level

University Level