June 30, 2026 - Calculus 2 - Finite and Infinite Series and Convergence Tests

Course Administration and Scheduling

  • Semester Progress: The semester is nearing its end. The current plan is to finalize the topic of series during this week.

  • Upcoming Examination:

    • A take-home test will be assigned on Thursday.

    • Due to student orientation conflicts on Thursday, the test will be posted online so physical attendance is not mandatory for its retrieval.

    • Deadline Policy: If the instructor finishes all intended material this week, the test will be due on Monday. If the material requires more time, the test will be assigned on Monday and due the following Thursday.

    • There will be no quiz on Thursday given the take-home test assignment.

Advanced Manipulation of Geometric Series

  • Refining the Geometric Form: To use the standard geometric series formula, the series must ideally be in the form ∑n=1∞arn−1\sum_{n=1}^{\infty} a r^{n-1}.

  • Index and Power Adjustment:

    • If a series index starts at n=3n=3 and the term is (23)n−1(\frac{2}{3})^{n-1}, one can shift the index to start at 11 by subtracting 2 from the index and adding 2 to the value of nn within the formula.

    • Algebraic Manipulation Example: To get a power of n+1n+1 down to n−1n-1 without changing the index:

      • Recognize that (23)n+1=(23)n×(23)1(\frac{2}{3})^{n+1} = (\frac{2}{3})^n \times (\frac{2}{3})^1.

      • Further, (23)n=(23)n−1×(23)1(\frac{2}{3})^n = (\frac{2}{3})^{n-1} \times (\frac{2}{3})^1.

      • By peeling off factors, you can restructure the expression into the arn−1a r^{n-1} format. For the example discussed, this involved factoring out (23)×(23)=49(\frac{2}{3}) \times (\frac{2}{3}) = \frac{4}{9}.

Geometric Series Identification and Calculation

  • Example 1: ∑n=1∞52n\sum_{n=1}^{\infty} \frac{5}{2^n}

    • Rewrite as 5∑(12)n5 \sum (\frac{1}{2})^n. Factor out a 12\frac{1}{2} to obtain the n−1n-1 exponent.

    • a=5a = 5, r=12r = \frac{1}{2}.

    • Sum calculation: a1−r=51−12=512=10\frac{a}{1-r} = \frac{5}{1 - \frac{1}{2}} = \frac{5}{\frac{1}{2}} = 10.

  • Example 2: ∑n=1∞−3n+14n−1\sum_{n=1}^{\infty} \frac{-3^{n+1}}{4^{n-1}}

    • The goal is to match the exponents of the numerator and denominator so they can be combined into a single ratio rr.

    • Numerator manipulation: −3n+1=−3(n−1)+2=(−32)×(−3n−1)=9×(−3)n−1-3^{n+1} = -3^{(n-1)+2} = (-3^2) \times (-3^{n-1}) = 9 \times (-3)^{n-1}. (Note: Care must be taken with the negative sign; the speaker notes a negative should be present in each step).

    • Combined form: ∑n=1∞9×(−3)n−14n−1=∑n=1∞9×(−34)n−1\sum_{n=1}^{\infty} \frac{9 \times (-3)^{n-1}}{4^{n-1}} = \sum_{n=1}^{\infty} 9 \times (\frac{-3}{4})^{n-1}.

    • Convergence Criterion: A geometric series converges if and only if |r| < 1. Since |-\frac{3}{4}| < 1, this series converges.

    • Sum calculation: a=9a = 9, r=−34r = -\frac{3}{4}. Sum = 91−(−34)=91+34=974=367\frac{9}{1 - (-\frac{3}{4})} = \frac{9}{1 + \frac{3}{4}} = \frac{9}{\frac{7}{4}} = \frac{36}{7}.

Telescoping Series

  • Definition: A telescoping series is a summation where the internal terms cancel each other out, leaving only a few terms at the beginning and the end.

  • Calculation via Partial Sums: The convergence of a telescoping series depends on the limit of the sequence of partial sums SkS_k.

  • Example (Partial Fraction Decomposition): ∑n=1∞1n(n+1)\sum_{n=1}^{\infty} \frac{1}{n(n+1)} decomposes into ∑n=1∞(1n−1n+1)\sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+1}).

    • Writing out terms: (1−12)+(12−13)+(13−14)+…(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots

    • The kk-th partial sum Sk=1−1k+1S_k = 1 - \frac{1}{k+1}.

    • Limit analysis: lim⁡k→∞(1−1k+1)=1−0=1\lim_{k \to \infty} (1 - \frac{1}{k+1}) = 1 - 0 = 1. The series converges to 1.

  • Trigonometric Example: ∑n=1∞(cos⁡(1n)−cos⁡(1n+1))\sum_{n=1}^{\infty} (\cos(\frac{1}{n}) - \cos(\frac{1}{n+1}))

    • Partial sum Sk=cos⁡(11)−cos⁡(1k+1)=cos⁡(1)−cos⁡(1k+1)S_k = \cos(\frac{1}{1}) - \cos(\frac{1}{k+1}) = \cos(1) - \cos(\frac{1}{k+1}).

    • Limit analysis: lim⁡k→∞(cos⁡(1)−cos⁡(1k+1))=cos⁡(1)−cos⁡(0)=cos⁡(1)−1\lim_{k \to \infty} (\cos(1) - \cos(\frac{1}{k+1})) = \cos(1) - \cos(0) = \cos(1) - 1.

The Divergence Test (n-th Term Test for Divergence)

  • Core Principle: For a series to converge, the terms being added (ana_n) must eventually approach zero. If the limit of the sequence ana_n is not zero, the sum is effectively adding non-negligible values infinitely many times, leading to infinity.

  • Formal Statement:

    • If lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0 or lim⁡n→∞an=±∞\lim_{n \to \infty} a_n = \pm \infty, then the series ∑an\sum a_n diverges.

  • Limitations (The Converse Trap): If lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the test is inconclusive. It does not guarantee convergence.

    • Classic Counter-example: The Harmonic Series ∑1n\sum \frac{1}{n}. Here, lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0, yet the series is known to diverge.

  • Applied Examples:

    • Example 1: ∑n=1∞n3n−1\sum_{n=1}^{\infty} \frac{n}{3n-1}. Limit as n→∞n \to \infty is 13\frac{1}{3}. Since 13≠0\frac{1}{3} \neq 0, the series diverges by the Divergence Test.

    • Example 2: ∑n=1∞cos⁡(1n)\sum_{n=1}^{\infty} \cos(\frac{1}{n}). Limit as n→∞n \to \infty is cos⁡(0)=1\cos(0) = 1. Since 1≠01 \neq 0, the series diverges.

    • Example 3: ∑sin⁡(1n)\sum \sin(\frac{1}{n}). Limit is sin⁡(0)=0\sin(0) = 0. The test is inconclusive.

    • Example 4: ∑e1n\sum e^{\frac{1}{n}}. Limit is e0=1e^0 = 1. Since 1≠01 \neq 0, the series diverges.

The Integral Test

  • Conceptual Basis: Comparing the sum of a series to the area under a curve. If a function f(x)f(x) is continuous, positive, and decreasing, then the behavior of the series ∑an\sum a_n (where an=f(n)a_n = f(n)) matches the behavior of the improper integral ∫N∞f(x)dx\int_{N}^{\infty} f(x) dx.

  • Criteria for Use:

    1. ana_n must be non-negative.

    2. There must exist a function f(x)f(x) such that f(n)=anf(n) = a_n.

    3. f(x)f(x) must be continuous.

    4. f(x)f(x) must be decreasing for all x≥Nx \ge N.

  • Relationship to Harmonic Series: Using the integral ∫1∞1xdx=[ln⁡(x)]1∞=∞\int_{1}^{\infty} \frac{1}{x} dx = [\ln(x)]_{1}^{\infty} = \infty, we prove the Harmonic Series diverges.

  • Example 1: ∑e−n\sum e^{-n}. Integral ∫1∞e−xdx=lim⁡t→∞[−e−x]1t=lim⁡t→∞(−e−t+e−1)=1e\int_{1}^{\infty} e^{-x} dx = \lim_{t \to \infty} [-e^{-x}]_{1}^{t} = \lim_{t \to \infty} (-e^{-t} + e^{-1}) = \frac{1}{e}.

    • The integral converges, therefore the series converges by the Integral Test.

    • Note: The value of the integral (1e\frac{1}{e}) is NOT the sum of the series. The sum of this geometric series is actually 1/e1−1/e=1e−1\frac{1/e}{1 - 1/e} = \frac{1}{e-1}.

  • Example 2: ∑1n3\sum \frac{1}{n^3}. Integral ∫1∞x−3dx=lim⁡t→∞[x−2−2]1t=lim⁡t→∞(−12t2+12)=12\int_{1}^{\infty} x^{-3} dx = \lim_{t \to \infty} [\frac{x^{-2}}{-2}]_{1}^{t} = \lim_{t \to \infty} (\frac{-1}{2t^2} + \frac{1}{2}) = \frac{1}{2}. The series converges.

  • Example 3: ∑12n−1\sum \frac{1}{\sqrt{2n-1}}.

    • Verify decreasing status: \frac{1}{\sqrt{2(n+1)-1}} < \frac{1}{\sqrt{2n-1}} simplifies to \sqrt{2n+1} > \sqrt{2n-1}, which is true.

    • Integral: ∫1∞(2x−1)−12dx\int_{1}^{\infty} (2x-1)^{-\frac{1}{2}} dx. Using u-substitution (u=2x−1,du=2dxu = 2x-1, du = 2dx), we get [2x−1]1∞=∞[\sqrt{2x-1}]_{1}^{\infty} = \infty. The series diverges.

P-Series Test

  • Definition: A series of the form ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}, where pp is a real number.

  • Convergence Rules: derived from the integral test on x−px^{-p}.

    • Converges if p > 1.

    • Diverges if p≤1p \le 1.

  • Applied Examples:

    • Example 1: ∑1n2\sum \frac{1}{n^2}. Since p = 2 > 1, it converges.

    • Example 2: ∑1n3=∑1n3/2\sum \frac{\sqrt{1}}{\sqrt{n^3}} = \sum \frac{1}{n^{3/2}}. Since p = \frac{3}{2} > 1, it converges.

    • Example 3: ∑1n45=∑1n4/5\sum \frac{1}{\sqrt[5]{n^4}} = \sum \frac{1}{n^{4/5}}. Since p=45≤1p = \frac{4}{5} \le 1, it diverges.

Direct Comparison Test

  • Concept: Comparing a "messy" series to a "clean" series (usually a P-series or geometric series) to determine behavior.

  • Rules:

    • If 0≤an≤bn0 \le a_n \le b_n and ∑bn\sum b_n converges, then ∑an\sum a_n converges (the smaller of two convergent terms must converge).

    • If 0≤bn≤an0 \le b_n \le a_n and ∑bn\sum b_n diverges, then ∑an\sum a_n diverges (the larger of two divergent terms must diverge).

  • Example Walkthrough: Comparing ∑1n2+1\sum \frac{1}{n^2+1} to ∑1n2\sum \frac{1}{n^2}.

    • The terms of 1n2+1\frac{1}{n^2+1} are always smaller than 1n2\frac{1}{n^2}.

    • Since ∑1n2\sum \frac{1}{n^2} is a convergent P-series (p=2p=2), the smaller series ∑1n2+1\sum \frac{1}{n^2+1} also converges.

Questions & Discussion

  • Question regarding Divergence Test logic: "So, what does that mean exactly? Like, is it saying that it's getting close to one, but it never actually meets?"

    • Response: If the limit is 1/31/3, it means as we get further and further in the series, we are essentially adding 1/31/3 over and over. Adding a constant value infinitely many times causes the sum to grow infinitely large.

  • Question regarding the Cosine limit: "We did the cosine of 1/n1/n and we got the limit. It wasn't like the same… I kept plugging in numbers and it just kept being closer to one but it wasn't like growing by one the same way."

    • Response: The function cos⁡(1/n)\cos(1/n) behaves differently than a linear growth. The input to the cosine (1/n1/n) is heading toward zero, which makes the output of the cosine head toward 1. There is a difference between the input behavior and the actual output (the term of the series).

  • Question on Limit values: "So then if it's greater than one then it diverges. If it's zero, it's inconclusive. If it's less than zero… then it's also divergent?"

    • Response: Any limit value other than zero—positive, negative, or infinite—indicates divergence. If the limit is zero, the test provides no information.