Math Problem Statement

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

Solution

The problem you provided asks to determine whether the series

n=1n(x4)n3n\sum_{n=1}^{\infty} \frac{n(x-4)^n}{3^n}

is absolutely convergent or divergent using the ratio test.

Solution

To apply the ratio test, we consider the absolute value of the general term of the series, an=nx4n3na_n = \frac{n |x-4|^n}{3^n}.

  1. Ratio Test Setup: The ratio test requires calculating the limit: L=limnan+1anL = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| where an=nx4n3na_n = \frac{n |x-4|^n}{3^n}.

  2. Compute an+1a_{n+1}: an+1=(n+1)x4n+13n+1=(n+1)x4n+133na_{n+1} = \frac{(n+1) |x-4|^{n+1}}{3^{n+1}} = \frac{(n+1) |x-4|^{n+1}}{3 \cdot 3^n}

  3. Set Up the Ratio: an+1an=(n+1)x4n+133n3nnx4n=(n+1)x43n\frac{a_{n+1}}{a_n} = \frac{(n+1) |x-4|^{n+1}}{3 \cdot 3^n} \cdot \frac{3^n}{n |x-4|^n} = \frac{(n+1) |x-4|}{3n}

  4. Take the Limit as nn \to \infty: L=limn(n+1)x43n=x43limnn+1n=x43L = \lim_{n \to \infty} \frac{(n+1) |x-4|}{3n} = \frac{|x-4|}{3} \lim_{n \to \infty} \frac{n+1}{n} = \frac{|x-4|}{3}

  5. Apply the Ratio Test Conclusion: The ratio test states that:

    • If L<1L < 1, the series converges absolutely.
    • If L>1L > 1, the series diverges.
    • If L=1L = 1, the test is inconclusive.

    Therefore, the series will converge absolutely if x43<1\frac{|x-4|}{3} < 1 which simplifies to x4<3|x-4| < 3

  6. Solution in Terms of xx: This inequality implies: 3<x4<3-3 < x - 4 < 3 or 1<x<71 < x < 7

Conclusion

The series converges absolutely for xx in the interval (1,7)(1, 7).

Would you like further details or have any questions?


Here are 5 related questions to expand your understanding:

  1. How would the ratio test differ if the series included a factorial in the numerator or denominator?
  2. Can the ratio test determine convergence if L=1L = 1?
  3. How does absolute convergence differ from conditional convergence?
  4. What would happen if the series was n=1n2(x4)n3n\sum_{n=1}^{\infty} \frac{n^2 (x-4)^n}{3^n}?
  5. Why is it helpful to check for absolute convergence when dealing with power series?

Tip: The ratio test is particularly useful for series with terms involving powers, as it simplifies with ratios.

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Math Problem Analysis

Mathematical Concepts

Series
Absolute Convergence
Ratio Test

Formulas

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

Theorems

Ratio Test

Suitable Grade Level

Grades 11-12