Math Problem Statement

Исследовать на абсолютную или условную сходимость: (a) ∑_{n=1}^∞ (-1)^n / (3n-2)! (b) ∑_{n=1}^∞ (-1)^n n(3n^2 - 1) / (5 + 2n^2)

Solution

The task involves determining whether the given series converge absolutely, conditionally, or diverge. Let's analyze the two series one by one.


Problem (a):

n=1(1)n(3n2)!\sum_{n=1}^\infty \frac{(-1)^n}{(3n-2)!}

Analysis:

  1. Absolute Convergence:
    To check for absolute convergence, consider the series of absolute values: n=11(3n2)!.\sum_{n=1}^\infty \frac{1}{(3n-2)!}. The factorial (3n2)!(3n-2)! grows extremely quickly, and we know that series of the form 1n!\sum \frac{1}{n!} converge because the terms tend to zero extremely fast. Thus, this series converges absolutely.

  2. Conditional Convergence:
    Since the series already converges absolutely, it automatically converges conditionally as well.


Problem (b):

n=1(1)nn(3n21)5+2n2.\sum_{n=1}^\infty (-1)^n \frac{n(3n^2 - 1)}{5 + 2n^2}.

Analysis:

  1. Absolute Convergence:
    Consider the series of absolute values: n=1n(3n21)5+2n2.\sum_{n=1}^\infty \left| \frac{n(3n^2 - 1)}{5 + 2n^2} \right|. For large nn, the numerator behaves like 3n33n^3 (dominant term), and the denominator behaves like 2n22n^2, so the terms simplify asymptotically to: n(3n2)2n23n2.\frac{n(3n^2)}{2n^2} \sim \frac{3n}{2}. Since 3n2\frac{3n}{2} grows without bound, the terms do not tend to zero, and the series diverges absolutely.

  2. Conditional Convergence:
    Now consider the original alternating series. For alternating series to converge, the terms must:

    • Decrease monotonically (an+1<ana_{n+1} < a_n).
    • Tend to zero (limnan=0\lim_{n \to \infty} a_n = 0).

    Since the absolute value of the terms grows like 3n2\frac{3n}{2}, they do not tend to zero. Therefore, the series does not converge conditionally either.


Final Results:

  • (a): The series converges absolutely.
  • (b): The series diverges.

Do you want a step-by-step explanation for any specific part? Let me know!


Related Questions:

  1. How does the factorial growth in series (a) ensure convergence?
  2. Why does the alternating series test fail in part (b)?
  3. Can series with terms tending to infinity ever converge conditionally?
  4. How can we test for divergence in non-alternating series?
  5. What are the differences between absolute and conditional convergence?

Tip:

When working with factorial-based terms, always compare their growth rate to standard tests (e.g., ratio test, comparison with ene^n). This simplifies deciding convergence behavior.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Absolute Convergence
Conditional Convergence
Factorial Growth
Alternating Series

Formulas

|∑ (-1)^n / (3n-2)!|
|∑ n(3n^2 - 1) / (5 + 2n^2)|

Theorems

Absolute Convergence Test
Alternating Series Test

Suitable Grade Level

Undergraduate Level