Calculus II: The Integral Test for Infinite Series Convergence

Administrative Information and Course Logistics

• Weekly Schedule and Deadlines:

  • Homework is due on Thursday.
  • There is no quiz on Thursday.
  • Attendance at discussion section is strongly advised regardless of the lack of a quiz, as active problem solving is essential for mastering mathematics and teaching assistants provide dedicated support.
  • Module 2 is due on Sunday.

• Exam Performance and Statistics:

  • Exam grades are released via Gradescope in the afternoon.
  • The exam median score is approximately $86$.
  • The exam mean score is slightly lower than the median, located in the mid-$85\text{s}$.

• Regrade Policy and Procedures:

  • Regrade requests open on Gradescope immediately after grades are published.
  • Students who identify rubric discrepancies (such as losing points despite correctly stating an expression like $x = 3\sin(x)$) may submit a formal request for the grader to review the submission.
  • All regrade communication must remain professional, kind, and polite toward the human grader, clearly articulating the mathematical justification for why credit should be awarded.

Introduction to Infinite Series Convergence

• The Nature of Series in Chapter 11:

  • For the vast majority of infinite series, determining the exact value to which the series converges is impossible with standard tools.
  • The primary objective in Chapter 11 is to answer the binary question: does the series converge, or does the series diverge?
  • Chapter 11 introduces a comprehensive suite of convergence tests designed to definitively determine convergence or divergence without necessarily finding the sum.

• Inventory of Known Convergence Tools Prior to the Integral Test:

  • Geometric series
  • Harmonic series
  • Alternating series
  • Divergence test (Test for Divergence)

The Integral Test: Definitions, Conditions, and Framework

• Fundamental Concept:

  • The Integral Test provides a direct connection between the convergence of an infinite series and the convergence of an improper integral.
  • Given an infinite series $\sum_{n=1}^{\infty} a_n$, an associated continuous real-valued function $f(x)$ is constructed such that $f(n) = a_n$ at all integer points $n$.
  • The transformation from a discrete series term to a continuous function typically involves replacing the discrete index variable $n$ with the continuous real variable $x$.

• The Three Mandatory Hypotheses:

  • For the Integral Test to be valid, the corresponding function $f(x)$ on the interval of interest (typically $[1, \infty)$) must satisfy three conditions:
  1. Continuous: The function must have no discontinuities, vertical asymptotes, or undefined values on the integration interval.
  2. Positive: $f(x) > 0$ across the entire integration interval.
  3. Decreasing: The function values must decrease as $x$ increases.

• Rigorous Theorem Statement:

  • Let $f(x)$ be a continuous, positive, and decreasing function on $[1, \infty)$, and let $a_n = f(n)$.
  • If the improper integral $\int_{1}^{\infty} f(x)\,dx$ converges, then the series $\sum_{n=1}^{\infty} a_n$ converges.
  • If the improper integral $\int_{1}^{\infty} f(x)\,dx$ diverges, then the series $\sum_{n=1}^{\infty} a_n$ diverges.

• Key Strengths and Structural Properties of the Test:

  • Decisiveness: Unlike certain tests that can yield inconclusive results, the Integral Test always produces a definitive conclusion (convergence or divergence), provided that the function satisfies the hypotheses and an antiderivative can be evaluated.
  • Three-Part Analytical Protocol: Every application of the Integral Test must formally include:
  1. Explicit statement and verification of the three hypotheses (continuous, positive, decreasing).
  2. Execution of the improper integral using formal limit notation.
  3. A clear conclusion explicitly naming the Integral Test and stating whether the original infinite series converges or diverges.

• Critical Caveats and Flexibility Rules:

  • Non-Unit Starting Indices:
  • The lower index does not need to be $1$.
  • If a series begins at any arbitrary positive integer $a$ (such as $\sum_{n=2}^{\infty} a_n$, $\sum_{n=5}^{\infty} a_n$, or $\sum_{n=15}^{\infty} a_n$), the Integral Test applies directly using the corresponding lower limit of integration: $\int_{a}^{\infty} f(x)\,dx$.
  • Eventual Decreasing Requirement:
  • A function does not need to decrease across its entire initial domain; it only needs to be eventually decreasing.
  • If a function increases initially over a finite interval and begins decreasing permanently past some critical threshold $x = c$, the Integral Test remains fully valid.
  • The initial non-decreasing portion represents a finite region with a finite, positive area that cannot cause an integral to diverge.
  • Oscillating functions that alternate infinitely between increasing and decreasing (such as $\sin(x)$) fail this condition and cannot be evaluated with the Integral Test.
  • Sum Value vs. Integral Value:
  • The numerical value of the improper integral does not equal the sum of the series.
  • If $\int_{1}^{\infty} f(x)\,dx = L$, the series converges, but its sum is generally not equal to $L$.

Fully Worked Example: Inverse Tangent Series

• Problem Statement:

  • Determine whether the series $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converges or diverges.

• Step 1: Function Definition:

  • Define the continuous real function: $f(x) = \frac{1}{x^2 + 1}$.

• Step 2: Verification of Conditions:

  • Continuity: The denominator $x^2 + 1$ is strictly positive for all real $x$, having no real zeros ($x^2 + 1 \ge 1$). Thus, $f(x)$ is continuous on $[1, \infty)$.
  • Positivity: For all $x \ge 1$, the numerator is $1 > 0$ and the denominator $x^2 + 1 > 0$. Hence, $f(x) > 0$.
  • Decreasing: The numerator is a constant ($1$), while the denominator $x^2 + 1$ increases monotonically as $x$ increases for all $x \ge 1$. Because the denominator grows larger while the numerator remains fixed, the fraction decreases. Thus, $f(x)$ is decreasing.

• Step 3: Setup and Evaluation of the Improper Integral:

  • Express the improper integral using formal limit notation:     $\int_{1}^{\infty} \frac{1}{x^2 + 1}\,dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^2 + 1}\,dx$
  • Recognize the standard antiderivative:     $\int \frac{1}{x^2 + 1}\,dx = \arctan(x)$
  • Apply the Fundamental Theorem of Calculus:     $\lim_{t \to \infty} \left[ \arctan(x) \right]_{1}^{t} = \lim_{t \to \infty} \left( \arctan(t) - \arctan(1) \right)$
  • Evaluate the individual limits and trigonometric values:
  • As $t \to \infty$, $\arctan(t) \to \frac{\pi}{2}$.
  • For $\arctan(1)$, find the angle whose tangent is $1$: $\arctan(1) = \frac{\pi}{4}$.
  • Compute the result:     $\lim_{t \to \infty} \left( \arctan(t) - \arctan(1) \right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

• Step 4: Formal Conclusion:

  • Because the improper integral converges to $\frac{\pi}{4}$, by the Integral Test, the series $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converges.

Methodological Rules for Substitution and Justification

• Justifying Decreasing Behavior:

  • When the variable appears exclusively in the denominator and the numerator is a constant, stating that the denominator increases is sufficient to establish that the function decreases.
  • When the variable appears in both the numerator and the denominator, verbal or intuitive arguments regarding relative growth rates are insufficient; explicit algebraic justification via the First Derivative Test ($f'(x) < 0$) is required.

• Rigorous Handling of $u$-Substitution in Improper Integrals:

  • When performing a $u$-substitution within an improper integral, mixing $x$-bounds with a $u$-differential (e.g., writing $\lim_{t \to \infty} \int_{1}^{t} u\,du$ when $1$ and $t$ are bounds for $x$) is mathematically invalid.
  • Two acceptable procedural paths exist:
  1. Transform the Bounds: Convert the limits of integration directly into values of $u$ using the substitution equation before integrating.
  2. Indefinite Integral Route: Treat the antiderivative as a separate indefinite integral, compute the antiderivative in terms of $u$, add $+ C$, substitute back into the original variable $x$, and then evaluate the definite integral across the original bounds using limit notation.

• Integration Mechanics of Linear Compositions:

  • In substitutions involving an expression of the form $u = ax + b$, the differential $du = a\,dx$ introduces a scaling factor of $\frac{1}{a}$.
  • Integrating terms such as $u^{-4}$ yields $\frac{u^{-3}}{-3}$, introducing a denominator factor of $-3$.
  • Combining a differential factor of $\frac{1}{3}$ and an integration factor of $\frac{1}{-3}$ produces an overall coefficient denominator of $9$ (specifically $-\frac{1}{9}$).

Advanced Worked Example: First Derivative Test and Logarithmic Series

• Problem Statement:

  • Determine whether the series $\sum_{n=1}^{\infty} \frac{\ln(n)}{n}$ converges or diverges.

• Step 1: Function Definition:

  • Define the continuous real function: $f(x) = \frac{\ln(x)}{x}$.

• Step 2: Verification of Conditions:

  • Continuity: The natural logarithm $\ln(x)$ is continuous for all $x > 0$, and the rational division by $x$ is continuous for all $x \neq 0$. Because the domain is $[1, \infty)$, $f(x)$ is continuous.
  • Positivity: For all $x > 1$, $\ln(x) > 0$ and $x > 0$, which means $f(x) > 0$. (At the initial point $x = 1$, $f(1) = \frac{\ln(1)}{1} = 0$, but the terms are strictly positive for all $x > 1$).
  • Decreasing (First Derivative Analysis):
  • Because both the numerator and the denominator contain $x$, intuitive comparisons fail; the First Derivative Test must be applied.
  • Apply the Quotient Rule, where the low function is $x$ and the high function is $\ln(x)$:       $f'(x) = \frac{\text{low} \cdot d(\text{high}) - \text{high} \cdot d(\text{low})}{(\text{low})^2} = \frac{x \cdot \left(\frac{1}{x}\right) - \ln(x) \cdot 1}{x^2}$
  • Simplify the numerator:       $f'(x) = \frac{1 - \ln(x)}{x^2}$
  • Set $f'(x) < 0$ to find where the function is decreasing:       $\frac{1 - \ln(x)}{x^2} < 0$
  • Because the denominator $x^2$ is strictly positive for all $x \ge 1$, the sign of $f'(x)$ is determined entirely by the numerator:       $1 - \ln(x) < 0$$1 < \ln(x)$
  • Apply the exponential base $e$ to both sides:       $e^1 < e^{\ln(x)} \implies x > e$
  • Because $e \approx 2.718$, the function is strictly decreasing for all $x > e$.
  • The function satisfies the requirement of being eventually decreasing.

• Step 3: Setup and Evaluation of the Improper Integral:

  • Maintain the original lower limit of integration from the series ($x = 1$) rather than changing it to $e$:     $\int_{1}^{\infty} \frac{\ln(x)}{x}\,dx = \lim_{t \to \infty} \int_{1}^{t} \frac{\ln(x)}{x}\,dx$
  • Apply $u$-substitution:
  • Let $u = \ln(x)$.
  • Then $du = \frac{1}{x}\,dx$.
  • Transform the integration bounds from $x$ to $u$:
  • Lower bound: when $x = 1$, $u = \ln(1) = 0$.
  • Upper bound: when $x = t$, $u = \ln(t)$.
  • Rewrite the integral entirely in terms of $u$:     $\lim_{t \to \infty} \int_{0}^{\ln(t)} u\,du$
  • Evaluate the antiderivative:     $\lim_{t \to \infty} \left[ \frac{u^2}{2} \right]_{0}^{\ln(t)} = \lim_{t \to \infty} \left( \frac{(\ln(t))^2}{2} - \frac{0^2}{2} \right) = \lim_{t \to \infty} \frac{(\ln(t))^2}{2}$
  • Compute the limit as $t \to \infty$:
  • As $t \to \infty$, $\ln(t) \to \infty$, so $\frac{(\ln(t))^2}{2} \to \infty$.

• Step 4: Formal Conclusion:

  • The improper integral diverges to $\infty$.
  • By the Integral Test, the series $\sum_{n=1}^{\infty} \frac{\ln(n)}{n}$ diverges.

Questions and Discussion

• Exam Score Distributions:

  • Question: What was the class mean on the exam compared to the median?
  • Response: The median score was approximately $86$. The mean is consistently slightly lower than the median in these distributions, falling in the mid-$85\text{s}$. Full statistics are communicated formally via email.

• Alternative Testing Methods (Comparison Test vs. Integral Test):

  • Question (from MJ): Could a series like $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ be evaluated using a $p$-series test or a comparison test instead of evaluating an improper integral?
  • Response: Yes, multiple tests can often successfully evaluate the same series. Comparison tests (detailed in Section 11.4) allow series to be compared directly to known $p$-series. The Integral Test is chosen here as a foundational tool, but alternative valid tests will expand throughout the unit.

• Requirements for Demonstrating Decreasing Behavior:

  • Question (from Nina): What specific work must be shown to justify that a function is decreasing, and is it permissible to use the First Derivative Test?
  • Response: When an expression has a constant numerator and an increasing denominator, stating that the denominator grows larger is sufficient for justification. However, when both the numerator and denominator contain the independent variable, intuitive assertions about which part grows faster are not acceptable. The First Derivative Test must be performed algebraically by computing $f'(x)$, setting $f'(x) < 0$, and solving for the domain of $x$ where the inequality holds.

• Selecting Lower Bounds for Eventually Decreasing Integrals:

  • Question: Since the function $f(x) = \frac{\ln(x)}{x}$ only begins decreasing once $x > e$, should the lower bound of integration in the improper integral be adjusted from $1$ to $e$?
  • Response: No. The improper integral should always retain the original lower bound of the given series (here, from $1$ to $t$). The theoretical validity of the Integral Test requires only that the function be eventually decreasing; the integration bounds themselves remain aligned with the initial index of the summation.