July 02, 2026 - Calculus 2 - Study Notes on Infinite Series, Convergence Tests, and Absolute Convergence

Definition of Absolute and Conditional Convergence

• Absolute Convergence: A series is defined as absolutely convergent if the absolute value of the series converges. If $\\sum |a_n|$ converges, then the series is said to converge absolutely.

  • For series with only positive terms, absolute convergence is the same as normal convergence.
  • This distinction is primarily used in reference to alternating series.

• Conditional Convergence: A series is called conditionally convergent if the series itself converges, but the absolute value of the series diverges. Specifically, if $\\sum a_n$ converges but $\\sum |a_n|$ diverges, the series is conditionally convergent.

The Alternating Series Test (AST)

• This test is used specifically for alternating series (series where terms alternate signs). The general term is written as $(-1)^{n} a_n$ or $(-1)^{n+1} a_n$, where $a_n > 0$.
• Conditions for AST:
  1. The limit of the sequence terms as $n$ approaches infinity must be zero: $\\lim_{n \\to \\infty} a_n = 0$.
  2. The sequence terms must be decreasing: $a_{n+1} \\le a_n$ for all $n$.
• If both conditions are met, the alternating series is guaranteed to converge.

Detailed Example: The Alternating Harmonic Series

• Series: $\\sum_{n=1}^{\\infty} \\frac{(-1)^{n+1}}{n}$
• Testing for Convergence (AST):
  • Identify $a_n = \\frac{1}{n}$.
  • Condition 1: $\\lim_{n \\to \\infty} \\frac{1}{n} = 0$. (True).
  • Condition 2: We check if $a_{n+1} < a_n \\implies \\frac{1}{n+1} < \\frac{1}{n}$. Since $n+1 > n$, the inequality holds. (True).
  • Result: The alternating series converges.
• Testing for Absolute Convergence:
  • Take the absolute value: $\\sum |\\frac{(-1)^{n+1}}{n}| = \\sum \\frac{1}{n}$.
  • This is the Harmonic Series, which is a P-series with $p = 1$.
  • Since $p \\le 1$, the harmonic series diverges.
• Final Classification: Because the alternating version converges but the absolute version diverges, the Alternating Harmonic Series is conditionally convergent.

Example Analysis: $\\sum \\frac{(-1)^{n+1}}{3n+1}$

• Convergence Check (AST):
  1. Limit check: $\\lim_{n \\to \\infty} \\frac{1}{3n+1} = 0$.
  2. Decreasing check: Compare $a_{n+1}$ and $a_n$.
     • $\\frac{1}{3(n+1)+1} = \\frac{1}{3n+4}$
     • Compare $\\frac{1}{3n+4}$ vs $\\frac{1}{3n+1}$. Cross multiplying gives $3n+1 < 3n+4$, which is always true.
  • The alternating series converges.
• Absolute Convergence Check:
  • Examine $\\sum \\frac{1}{3n+1}$.
  • Use the Limit Comparison Test (LCT) against the Divergent Harmonic Series $\\sum \\frac{1}{n}$.
  • Limit Calculation: $\\lim_{n \\to \\infty} \\frac{1/(3n+1)}{1/n} = \\lim_{n \\to \\infty} \\frac{n}{3n+1}$.
  • Using L'H\u00f4pital's Rule or leading coefficients: $\\frac{d/dn(n)}{d/dn(3n+1)} = \\frac{1}{3}$.
  • Since the result is a finite number ($1/3$) that is not zero or infinity, both series behave the same. Therefore, $\\sum \\frac{1}{3n+1}$ diverges.
• Classification: Conditionally Convergent.

Example Analysis: $\\sum (-1)^n \\frac{2n^2+1}{n^3+1}$

• Convergence Check (AST):
  1. Limit check: $\\lim_{n \\to \\infty} \\frac{2n^2+1}{n^3+1} = 0$ (denominator degree is higher).
  2. Decreasing check: Show $a_{n+1} < a_n$.
     • Checking $\\frac{2(n+1)^2+1}{(n+1)^3+1} < \\frac{2n^2+1}{n^3+1}$.
     • After algebraic expansion and distribution, one finds the numerator contains higher powers of $n$ that eventually force the terms to decrease for $n \\ge 1$.
  • The alternating series converges.
• Absolute Convergence Check:
  • Compare $a_n = \\frac{2n^2+1}{n^3+1}$ to the P-series $\\sum \\frac{1}{n}$.
  • LCT: $\\lim_{n \\to \\infty} \\frac{(2n^2+1)/(n^3+1)}{1/n} = \\lim_{n \\to \\infty} \\frac{2n^3+n}{n^3+1} = 2$.
  • Since the limit is a non-zero constant ($2$) and $\\sum \\frac{1}{n}$ diverges, this series diverges absolutely.
  • Instead, if the original series was compared to $\\sum \\frac{1}{n^2}$:
    • LCT: $\\lim_{n \\to \\infty} \\frac{(2n^2+1)/(n^3+1)}{1/n^2} = \\lim_{n \\to \\infty} \\frac{2n^4+n^2}{n^3+1} = \\infty$.
    • This suggests the original choice of $\\frac{1}{n}$ was correct and the series is conditionally convergent.

The Ratio Test

• The Ratio Test is highly effective for series involving factorials or exponentials.
• Procedure: Calculate the limit $L = \\lim_{n \\to \\infty} \\left| \\frac{a_{n+1}}{a_n} \\right|$.
• Results:
  1. If $L < 1$: The series converges absolutely.
  2. If $L > 1$ or $L = \\infty$: The series diverges.
  3. If $L = 1$: The test is inconclusive (the series could be absolutely convergent, conditionally convergent, or divergent).
• Note on Absolute Value: The Ratio Test explicitly checks for absolute convergence because of the absolute value bars in the formula.

Factorial Properties and the Ratio Test

• Definitions and Identifiers:
  • $0! = 1$
  • $1! = 1$
  • $2! = 2 \\times 1$
  • $3! = 3 \\times 2 \\times 1$
  • General form: $n! = n \\times (n-1) \\times ... \\times 1$
• Key Simplification Shorthands:
  • $(n+1)! = (n+1) \\times n!$
  • $(n+2)! = (n+2)(n+1) \\times n!$
  • $(2n+2)! = (2n+2)(2n+1) \\times (2n)!$

Ratio Test Example: $\\sum \\frac{2^n}{n!}$

• Setup:
  • $a_n = \\frac{2^n}{n!}$
  • $a_{n+1} = \\frac{2^{n+1}}{(n+1)!}$
• Calculation:
  • $L = \\lim_{n \\to \\infty} \\left| \\frac{2^{n+1}}{(n+1)!} \\times \\frac{n!}{2^n} \\right|$
  • Simplify the powers of 2: $\\frac{2^{n+1}}{2^n} = 2$
  • Simplify the factorials: $\\frac{n!}{(n+1)!} = \\frac{n!}{(n+1)n!} = \\frac{1}{n+1}$
  • $L = \\lim_{n \\to \\infty} \\frac{2}{n+1} = 0$
• Result: Since $0 < 1$, the series converges absolutely.

Ratio Test Example: $\\sum \\frac{n^n}{n!}$

• Setup:
  • $a_{n+1} = \\frac{(n+1)^{n+1}}{(n+1)!}$
• Calculation:
  • $L = \\lim_{n \\to \\infty} \\left| \\frac{(n+1)^{n+1}}{(n+1)!} \\times \\frac{n!}{n^n} \\right|$
  • Simplify factorials: $\\frac{n!}{(n+1)!} = \\frac{1}{n+1}$
  • Current limit: $\\lim_{n \\to \\infty} \\frac{(n+1)^{n+1}}{(n+1)n^n} = \\lim_{n \\to \\infty} \\frac{(n+1)^n}{n^n}$
  • Rewrite as a single power: $\\lim_{n \\to \\infty} \\left( \\frac{n+1}{n} \\right)^n = \\lim_{n \\to \\infty} \\left( 1 + \\frac{1}{n} \\right)^n$
• Identification of Transcendental Number: This specific limit is the definition of $e$.
  • $e \\approx 2.718$
• Result: Since $e > 1$, the series diverges.

Ratio Test Example: $\\sum (-1)^n \\frac{(n!)^2}{(2n)!}$

• Setup:
  • $a_{n+1} = \\frac{((n+1)!)^2}{(2(n+1))!} = \\frac{((n+1)!)^2}{(2n+2)!}$
• Calculation:
  • $L = \\lim_{n \\to \\infty} \\left| \\frac{((n+1)!)^2}{(2n+2)!} \\times \\frac{(2n)!}{(n!)^2} \\right|$
  • Group terms: $\\left( \\frac{(n+1)!}{n!} \\right)^2 \\times \\frac{(2n)!}{(2n+2)!}$
  • Simplify square part: $\\left( \\frac{(n+1)n!}{n!} \\right)^2 = (n+1)^2$
  • Simplify factorial part: $\\frac{(2n)!}{(2n+2)(2n+1)(2n)!} = \\frac{1}{(2n+2)(2n+1)}$
  • Limit: $\\lim_{n \\to \\infty} \\frac{(n+1)^2}{(2n+2)(2n+1)} = \\lim_{n \\to \\infty} \\frac{n^2 + 2n + 1}{4n^2 + 6n + 2}$
  • The ratio of leading coefficients is $\\frac{1}{4}$.
• Result: Since $\\frac{1}{4} < 1$, the series converges absolutely.

The Root Test

• The Root Test is used primarily when terms are raised to the power of $n$.
• Procedure: Calculate the limit $L = \\lim_{n \\to \\infty} \\sqrt[n]{|a_n|} = \\lim_{n \\to \\infty} |a_n|^{1/n}$.
• Results:
  1. If $L < 1$: The series converges absolutely.
  2. If $L > 1$ or $L = \\infty$: The series diverges.
  3. If $L = 1$: The test is inconclusive.

Root Test Example 1: $\\sum \\left( \\frac{2n^2+1}{3n^2-1} \\right)^n$

• Calculation:
  • $L = \\lim_{n \\to \\infty} \\sqrt[n]{\\left| \\left( \\frac{2n^2+1}{3n^2-1} \\right)^n \\right|}$
  • The $n$-th power and $n$-th root cancel out.
  • $L = \\lim_{n \\to \\infty} \\frac{2n^2+1}{3n^2-1} = \\frac{2}{3}$
• Result: Since $\\frac{2}{3} < 1$, the series converges absolutely.

Root Test Example 2: $\\sum \\left( \\frac{\\ln(n)}{n} \\right)^n$

• Calculation:
  • $L = \\lim_{n \\to \\infty} \\sqrt[n]{\\left| \\left( \\frac{\\ln(n)}{n} \\right)^n \\right|}$
  • Wait for normalization: $\\lim_{n \\to \\infty} \\frac{\\ln(n)}{n}$.
  • By L'H\u00f4pital's Rule: $\\lim_{n \\to \\infty} \\frac{1/n}{1} = 0$.
• Result: Since $0 < 1$, the series converges absolutely.

Scenario Where Ratio/Root Tests Fail

• Test Case: Alternating Harmonic Series $\\frac{(-1)^{n+1}}{n}$.
  • Applying the Ratio Test:
    • $L = \\lim_{n \\to \\infty} \\left| \\frac{(-1)^{n+2}/(n+1)}{(-1)^{n+1}/n} \\right|$
    • $L = \\lim_{n \\to \\infty} \\frac{n}{n+1} = 1$.
  • Result: Inconclusive.
  • Explanation: The Ratio and Root tests only detect absolute convergence. They cannot distinguish between conditional convergence and divergence when $L=1$. Because the alternating harmonic series is conditionally convergent, these tests fail to provide helpful information.