July 02, 2026 - Calculus 2 - Series Convergence and Calculus Tests Flashcards

Absolute and Conditional Convergence

  • Absolute Convergence: A series ∑an\sum a_n is defined as absolutely convergent if the absolute value of the series, ∑∣an∣\sum |a_n|, converges. For series with only positive terms, taking the absolute value changes nothing; therefore, absolute convergence is primarily a distinction used for alternating series.

  • Conditional Convergence: A series is conditionally convergent if the series itself, ∑an\sum a_n, converges, but its absolute value, ∑∣an∣\sum |a_n|, diverges.

  • The Alternating Harmonic Series Example:

    • Series: ∑(−1)n+1n\sum \frac{(-1)^{n+1}}{n}.

    • The Alternating Series Test (AST) requires checking two conditions for the sequence ana_n (without the alternating part):

      1. The limit of the sequence is zero: lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0.

      2. The sequence is non-increasing (terms are getting smaller): \frac{1}{n+1} < \frac{1}{n} is always true for positive nn.

    • Since both conditions are met, the alternating harmonic series converges.

    • Testing for absolute convergence: ∑∣(−1)n+1n∣=∑1n\sum |\frac{(-1)^{n+1}}{n}| = \sum \frac{1}{n}. This is the harmonic series (a P-series with P=1P=1), which is divergent.

    • Conclusion: The alternating harmonic series is conditionally convergent.

Alternating Series Test Application and Testing Absolute Convergence

  • **Example: ** ∑(−1)n3n+1\sum \frac{(-1)^n}{3n+1}

    • Convergence Check (AST):

      • Condition 1: lim⁡n→∞13n+1=0\lim_{n \to \infty} \frac{1}{3n+1} = 0.

      • Condition 2: Compare the (n+1)(n+1) term to the nn term: \frac{1}{3(n+1)+1} < \frac{1}{3n+1} \rightarrow \frac{1}{3n+4} < \frac{1}{3n+1}. Cross-multiplying yields 3n+1 < 3n+4, which is true. Thus, the alternating series converges.

    • Absolute Convergence Check:

      • The absolute value series is ∑13n+1\sum \frac{1}{3n+1}.

      • Testing ∑13n+1\sum \frac{1}{3n+1} using the Limit Comparison Test (LCT) with the divergent harmonic series ∑1n\sum \frac{1}{n}.

      • Limit: lim⁡n→∞1/(3n+1)1/n=lim⁡n→∞n3n+1\lim_{n \to \infty} \frac{1/(3n+1)}{1/n} = \lim_{n \to \infty} \frac{n}{3n+1}.

      • Using L'Hôpital's Rule or comparing degrees: lim⁡n→∞13=1/3\lim_{n \to \infty} \frac{1}{3} = 1/3.

      • Since 1/31/3 is a finite, non-zero number, both series behave the same way. Thus, ∑13n+1\sum \frac{1}{3n+1} diverges.

    • Result: This series is conditionally convergent.

  • **Example: ** ∑(−1)n(2n2+1)n3+1\sum \frac{(-1)^n (2n^2+1)}{n^3+1}

    • Checking terms for AST: The degree of the numerator is higher relative to the P-series chosen previously. The speaker illustrates checking for absolute convergence by comparing degrees.

    • Comparing to a P-series where P=2P=2: If we compare a similar rational function 2n2n4\frac{2n^2}{n^4}, it reduces to 2/n22/n^2, which converges.

    • Conclusion Logic: If the absolute value series converges (because the degree of the denominator is at least two higher than the numerator), the series converges absolutely. If the absolute series diverges but the original passes AST, it is conditional.

The Ratio Test

  • The Ratio Test is highly effective for series involving factorials or combinations of exponentials and polynomials because it eliminates the need to provide a comparison series.

  • Test Definition: Let L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|

    1. If L < 1, the series is absolutely convergent.

    2. If L > 1 or L=∞L = \infty, the series is divergent.

    3. If L=1L = 1, the test is inconclusive.

Working with Factorials

  • Factorial Basics:

    • 0!=10! = 1

    • 1!=11! = 1

    • 2!=2×1=22! = 2 \times 1 = 2

    • 3!=3×2×1=63! = 3 \times 2 \times 1 = 6

  • General Rules:

    • n!=n×(n−1)×(n−2)×⋯×1n! = n \times (n-1) \times (n-2) \times \dots \times 1

    • An expression can be rewritten as a lower factorial: (n+1)!=(n+1)×n!(n+1)! = (n+1) \times n!

    • This property allows for critical cancellations in the limit of the Ratio Test: n!(n+1)!=n!(n+1)n!=1n+1\frac{n!}{(n+1)!} = \frac{n!}{(n+1)n!} = \frac{1}{n+1}.

Ratio Test Examples

  • **Example 1: ** ∑2nn!\sum \frac{2^n}{n!}

    • Set up the ratio: lim⁡n→∞∣2n+1(n+1)!×n!2n∣\lim_{n \to \infty} |\frac{2^{n+1}}{(n+1)!} \times \frac{n!}{2^n}|

    • Simplify terms: lim⁡n→∞2n+12n×n!(n+1)!=lim⁡n→∞2n+1\lim_{n \to \infty} \frac{2^{n+1}}{2^n} \times \frac{n!}{(n+1)!} = \lim_{n \to \infty} \frac{2}{n+1}.

    • As n→∞n \to \infty, the limit is 00.

    • Since 0 < 1, the series converges absolutely.

  • **Example 2: ** ∑nnn!\sum \frac{n^n}{n!}

    • Set up the ratio: lim⁡n→∞∣(n+1)n+1(n+1)!×n!nn∣\lim_{n \to \infty} |\frac{(n+1)^{n+1}}{(n+1)!} \times \frac{n!}{n^n}|

    • Simplify: lim⁡n→∞(n+1)(n+1)n(n+1)n!×n!nn=lim⁡n→∞(n+1)nnn\lim_{n \to \infty} \frac{(n+1)(n+1)^n}{(n+1)n!} \times \frac{n!}{n^n} = \lim_{n \to \infty} \frac{(n+1)^n}{n^n}.

    • Rewrite as a single base: lim⁡n→∞(n+1n)n=lim⁡n→∞(1+1n)n\lim_{n \to \infty} (\frac{n+1}{n})^n = \lim_{n \to \infty} (1+\frac{1}{n})^n.

    • Famous Limit: The limit of (1+1/n)n(1+1/n)^n as n→∞n \to \infty is the value ee (approximately 2.7182.718).

    • Since e > 1, the series diverges according to the Ratio Test.

  • **Example 3: ** ∑(−1)n(n!)2(2n)!\sum \frac{(-1)^n (n!)^2}{(2n)!}

    • Set up ratio: lim⁡n→∞∣((n+1)!)2(2(n+1))!×(2n)!(n!)2∣\lim_{n \to \infty} |\frac{((n+1)!)^2}{(2(n+1))!} \times \frac{(2n)!}{(n!)^2}|

    • Expand (2(n+1))!(2(n+1))! as (2n+2)!=(2n+2)(2n+1)(2n)!(2n+2)! = (2n+2)(2n+1)(2n)!

    • Expand ((n+1)!)2((n+1)!)^2 as (n+1)2(n!)2(n+1)^2 (n!)^2

    • Simplification: (n+1)2(n!)2(2n+2)(2n+1)(2n)!×(2n)!(n!)2=(n+1)2(2n+2)(2n+1)\frac{(n+1)^2 (n!)^2}{(2n+2)(2n+1)(2n)!} \times \frac{(2n)!}{(n!)^2} = \frac{(n+1)^2}{(2n+2)(2n+1)}.

    • The leading polynomial terms are n24n2\frac{n^2}{4n^2}. The limit is 1/41/4.

    • Since 1/4 < 1, the series converges absolutely.

The Root Test

  • The Root Test is generally used for series where terms contain an exponent of nn.

  • Test Definition: Let L=lim⁡n→∞∣an∣nL = \lim_{n \to \infty} \sqrt[n]{|a_n|}

    1. If 0 \le L < 1, the series is absolutely convergent.

    2. If L > 1 or L=∞L = \infty, the series is divergent.

    3. If L=1L = 1, the test is inconclusive.

  • Example: Series involving terms like (2n2+13n2+1)n(\frac{2n^2+1}{3n^2+1})^n.

    • Apply the Root Test: ∣(2n2+13n2+1)n∣n=2n2+13n2+1\sqrt[n]{|(\frac{2n^2+1}{3n^2+1})^n|} = \frac{2n^2+1}{3n^2+1}.

    • Take the limit: lim⁡n→∞2n2+13n2+1=2/3\lim_{n \to \infty} \frac{2n^2+1}{3n^2+1} = 2/3.

    • Result: Since 2/3 < 1, the series converges absolutely.

Limitation: Inconclusive Tests

  • The Ratio and Root tests only detect absolute convergence. They cannot distinguish conditional convergence.

  • Alternating Harmonic Series via Ratio Test:

    • lim⁡n→∞∣1/(n+1)1/n∣=lim⁡n→∞nn+1=1\lim_{n \to \infty} |\frac{1/(n+1)}{1/n}| = \lim_{n \to \infty} \frac{n}{n+1} = 1.

    • The result L=1L=1 means the test is inconclusive. This happens because the series does not converge absolutely, yet it does not diverge; it resides in the state of conditional convergence which these specific tests do not identify.

Questions & Discussion

  • Question: Is it the same result whether the index shift is (n−1)(n-1) or (n+1)(n+1)?

  • Answer: Yes; technically, one just shifts the series over by a term. When dealing with factorials, you decrement from the highest term down to the factorial you want to cancel (e.g., (2n+2)!(2n+2)! goes down to (2n)!(2n)!).

  • Question: Can you just divide the 22 out of the coefficient in (2n)!(2n)!?

  • Answer: No. (2n)!(2n)! means taking the value of 2n2n and multiplying it by every integer less than it until 11. It does not mean every term in the factorial is divisible by 22.

  • Student Roster and Attendance: The instructor called the following names during the take-home test distribution:

    • Matthew (is Ethan)

    • Gavin Davis

    • Owen (is also Alan Celeste)

    • Antonio

    • Elena (and person)

    • Lindsay

    • Belle (also passive Dylan)