Math Problem Statement

sigma n=1 (-1)^n+3 n-1/n^2+n convergence or divergence

Solution

To determine whether the given series

n=1(1)n+3n1n2+n\sum_{n=1}^\infty \frac{(-1)^n + 3n - 1}{n^2 + n}

converges or diverges, we analyze the terms step by step.

Step 1: Examine the general term

The general term of the series is:

an=(1)n+3n1n2+n.a_n = \frac{(-1)^n + 3n - 1}{n^2 + n}.

Here, (1)n(-1)^n introduces an alternating component, while 3n13n - 1 grows linearly as nn increases.

Step 2: Check the dominant term as nn \to \infty

As nn becomes large, the denominator n2+nn^2 + n grows quadratically, while the numerator grows linearly due to the 3n3n term. Thus, the general term ana_n behaves like:

an3nn2+n=3n.a_n \sim \frac{3n}{n^2 + n} = \frac{3}{n}.

This suggests the series may behave similarly to the harmonic series, which diverges.

Step 3: Apply the divergence test

The divergence test (or nn-th term test) states that if limnan0\lim_{n \to \infty} a_n \neq 0, the series diverges. Let's compute the limit of ana_n:

limnan=limn(1)n+3n1n2+n.\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{(-1)^n + 3n - 1}{n^2 + n}.

Focus on the dominant terms in the numerator and denominator. As nn \to \infty, the 3n3n term in the numerator dominates (1)n(-1)^n and 1-1, and the n2n^2 term in the denominator dominates nn. Thus:

an3nn2=3n.a_n \sim \frac{3n}{n^2} = \frac{3}{n}.

Since 3n0\frac{3}{n} \to 0 as nn \to \infty, the divergence test is inconclusive. We proceed to further tests.

Step 4: Analyze the series for convergence

The series is neither purely alternating nor purely positive, so direct tests for alternating or positive series do not apply straightforwardly. However, due to the growth rate of the terms, we suspect divergence. The comparison test or limit comparison test with a divergent series such as the harmonic series 1n\sum \frac{1}{n} might be useful.

Comparison with 1n\sum \frac{1}{n}:

an=(1)n+3n1n2+n3nn2+n=3n+1.a_n = \frac{(-1)^n + 3n - 1}{n^2 + n} \geq \frac{3n}{n^2 + n} = \frac{3}{n + 1}.

Since 3n+1\sum \frac{3}{n + 1} diverges (as it behaves like a harmonic series), the original series also diverges by comparison.


Conclusion:

The series n=1(1)n+3n1n2+n\sum_{n=1}^\infty \frac{(-1)^n + 3n - 1}{n^2 + n} diverges.


Do you want a detailed explanation of any of these steps?

Related Questions:

  1. How is the divergence test applied in other contexts?
  2. What is the comparison test and when is it used?
  3. How does alternating behavior affect convergence tests?
  4. Why does the harmonic series diverge?
  5. What are some examples of series that look similar but converge?

Tip:

Always simplify the dominant terms in the numerator and denominator to approximate the behavior of the series at infinity.

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Math Problem Analysis

Mathematical Concepts

Series
Convergence and Divergence
Asymptotic Analysis
Comparison Test
Divergence Test

Formulas

General term of the series: a_n = ((-1)^n + 3n - 1) / (n^2 + n)
Divergence test: If lim (n → ∞) a_n ≠ 0, the series diverges
Comparison with harmonic series: a_n ≈ 3/n

Theorems

Divergence Test (n-th Term Test)
Comparison Test

Suitable Grade Level

Undergraduate (Calculus II)