Lecture 3 Notes – Convergence Tests Deep-Dive

Course Logistics & Resources

  • Practice materials now on course website

    • 2 practice exams uploaded

    • 1 is an actual past mid-semester exam (same length, topics & difficulty as upcoming test)

    • 1 additional mock exam

    • Extra weekly problem sets

    • Scans of textbook problems + some solutions

    • Full solutions will NOT be posted for every problem, but help is available:

      • Lecturer office hours

      • Maths Drop-in Centre

      • Demonstrator consultation sessions

  • Timeline reminders

    • We are at the start of Week 3

    • Mid-semester exam: Monday of Week 6

    • Assignment 1 (≈ 4–5 multi-part questions)

    • Released end of this week / start of next

    • Due some time in Week 7 (after the mid-semester exam)

Toolbox of Series Convergence Tests (big picture)

  • Previously covered: p-series, comparison, integral test

  • Today’s focus (mostly examples):

    • Finishing Integral-Test examples

    • Divergence Test (a.k.a. “nth-term test”)

    • Direct & Limit Comparison

    • Continuous–function limit trick

    • Alternating-Series Test

    • Conditional vs Absolute convergence

    • Ratio Test

    • Root Test

Integral Test Example — ∑k=1∞ke−k\sum_{k=1}^{\infty} k e^{-k}

  • Analogous function: f(x)=xe−xf(x)=x e^{-x} (positive, decreasing for x≥1x\ge1)

  • Evaluate improper integral ∫1∞xe−x dx\int_{1}^{\infty} x e^{-x}\,dx

    • Integration by parts

    • Let u=x ,  dv=e−xdx  ⇒  du=dx ,  v=−e−xu=x\,,\;dv=e^{-x}dx\;\Rightarrow\;du=dx\,,\;v=-e^{-x}

    • After computation:
      ∫<em>1∞xe−xdx=[−xe−x−e−x]</em>1∞=2e\int<em>{1}^{\infty}x e^{-x}dx=\Bigl[-x e^{-x}-e^{-x}\Bigr]</em>{1}^{\infty}=\frac{2}{e}

    • Finite ⇒ series converges (by Integral Test).

“Continuous–Function Commutes with Limit” Trick

  • Fact: If ff is continuous and lim⁡<em>n→∞a</em>n=L\lim<em>{n\to\infty} a</em>n=L, then lim⁡<em>n→∞f(a</em>n)=f(L)\lim<em>{n\to\infty} f(a</em>n)=f(L).

  • Example series: ∑n=1∞n2n2+20n+93\sum_{n=1}^{\infty}\sqrt[3]{\dfrac{n^2}{n^2+20n+9}}

    • Treat cube-root as continuous f(x)=x1/3f(x)=x^{1/3}

    • First find inner limit:
      lim⁡n→∞n2n2+20n+9=1\lim_{n\to\infty}\frac{n^2}{n^2+20n+9}=1

    • Therefore overall term →f(1)=1\to f(1)=1 (≠ 0) ⇒ fails Divergence Test ⇒ series diverges.

    • Key algebra: divide top & bottom by n2n^2 to reveal dominant terms.

Divergence (“nth-Term”) Test Refresher

  • If lim⁡<em>n→∞a</em>n≠0\displaystyle\lim<em>{n\to\infty}a</em>n\neq0, then ∑an\sum a_n diverges — no further work needed.

  • Often fastest route to show divergence (called the “cheap” method in class).

Comparison & Limit-Comparison Examples

  • Target series: ∑n=1∞11+n2\sum_{n=1}^{\infty}\dfrac{1}{1+n^2}

    • Looks like 1n2\dfrac{1}{n^2} (a p-series with p=2>1p=2>1 → convergent)

  • Three possible strategies:

    1. Direct comparison: 11+n2≤1n2\dfrac{1}{1+n^2}\le\dfrac{1}{n^2} ⇒ convergent.

    2. Limit Comparison (quickest):
      lim⁡<em>n→∞11+n21n2=lim⁡</em>n→∞n21+n2=1\lim<em>{n\to\infty}\frac{\frac{1}{1+n^2}}{\frac{1}{n^2}}=\lim</em>{n\to\infty}\frac{n^2}{1+n^2}=1 (finite ≠0) ⇒ same behaviour as ∑1n2\sum\frac{1}{n^2} ⇒ convergent.

    3. Integral Test also works but is longer.

Alternating-Series Test (Leibniz Criterion)

  • Form: ∑<em>n=1∞(−1)n−1b</em>n\sum<em>{n=1}^{\infty}(-1)^{n-1} b</em>n with bn≥0b_n\ge0.

  • Converges if

    1. b<em>n+1≤b</em>nb<em>{n+1}\le b</em>n (monotone decreasing)

    2. lim⁡<em>n→∞b</em>n=0\displaystyle\lim<em>{n\to\infty}b</em>n=0

  • Classic example: Alternating harmonic series
    ∑n=1∞(−1)n−11n=ln⁡2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}=\ln 2 (convergent even though harmonic series alone diverges).

Quick diagnostics on sample alternating series

Series

bnb_n behaviour

lim⁡bn\lim b_n

Verdict

∑(−1)n−1 3n4n−1\sum(-1)^{n-1}\,\dfrac{3n}{4n-1}

bnb_n decreasing

→34≠0\to\tfrac34\neq0

Divergent (fails limit)

∑(−1)n n2n2+1\sum(-1)^{n}\,\dfrac{n^2}{n^2+1}

bnb_n increasing, →1≠0\to1\neq0

Divergent


Alternating harmonic ∑(−1)n−11n\sum(-1)^{n-1}\dfrac1n

bnb_n ↓, →0\to0

Convergent


Absolute vs Conditional Convergence

  • Absolute convergence: ∑∣an∣\sum |a_n| converges.

  • Conditional: ∑a<em>n\sum a<em>n converges but ∑∣a</em>n∣\sum|a</em>n| diverges.

  • Hierarchy: Absolute ⇒ Conditional, proved via triangle inequality
    ∣∑<em>n=1Na</em>n∣≤∑<em>n=1N∣a</em>n∣\Bigl|\sum<em>{n=1}^N a</em>n\Bigr|\le \sum<em>{n=1}^N |a</em>n|.

Examples

  • an=(−1)n−11na_n=(-1)^{n-1}\dfrac{1}{n}: conditional (series converges, absolute diverges).

  • an=(−1)n−11n2a_n=(-1)^{n-1}\dfrac{1}{n^2}: absolutely (and hence conditionally) convergent.

  • an=(−1)n3n4n−1a_n=(-1)^n\dfrac{3n}{4n-1}: neither (fails absolute; fails alternating-series test).

Ratio Test

  • Gadget: R<em>n=∣a</em>n+1an∣→n→∞LR<em>n=\left|\dfrac{a</em>{n+1}}{a_n}\right|\xrightarrow[n\to\infty]{}L

    • If L<1L<1 ⇒ absolutely convergent

    • If L>1L>1 ⇒ divergent

    • If L=1L=1 ⇒ inconclusive

  • Think of LL as the “effective ratio” vs a geometric series ∑arn\sum ar^n.

Example 1

∑n=1∞(−1)nn33n\sum_{n=1}^{\infty} (-1)^{n}\frac{n^3}{3^n}

  • Compute
    L=lim⁡<em>n→∞∣(n+1)3/3n+1n3/3n∣=13lim⁡</em>n→∞(n+1n)3=13L=\lim<em>{n\to\infty}\left|\frac{(n+1)^3/3^{n+1}}{n^3/3^{n}}\right|=\frac13\lim</em>{n\to\infty}\left(\frac{n+1}{n}\right)^3=\frac13

  • L=13<1L=\tfrac13<1 ⇒ series absolutely converges.

Example 2 (factorials)

∑n=1∞nnn!\sum_{n=1}^{\infty}\frac{n^n}{n!}

  • Ratio
    L=lim⁡<em>n→∞(n+1)n+1/(n+1)!nn/n!=lim⁡</em>n→∞(n+1)nnn=lim⁡n→∞(1+1n)n=e>1L=\lim<em>{n\to\infty}\frac{(n+1)^{n+1}/(n+1)!}{n^n/n!}=\lim</em>{n\to\infty}\frac{(n+1)^{n}}{n^n}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e>1

  • Divergent.

  • Factorials & exponentials cancel neatly in ratios — ideal for this test.

Root Test

  • Gadget: ∣an∣1/n→n→∞L\left|a_n\right|^{1/n}\xrightarrow[n\to\infty]{}L

    • L<1L<1 ⇒ absolute convergence

    • L>1L>1 ⇒ divergence

    • L=1L=1 ⇒ inconclusive

Example

∑k=2∞(2k−1k2+3)!k\sum_{k=2}^{\infty}\left(\frac{2k-1}{k^2+3}\right)^{!k}

  • Apply root test:
    L=lim⁡<em>k→∞∣2k−1k2+3∣=lim⁡</em>k→∞2kk2=lim⁡k→∞2k=0<1L=\lim<em>{k\to\infty}\left|\frac{2k-1}{k^2+3}\right|=\lim</em>{k\to\infty}\frac{2k}{k^2}=\lim_{k\to\infty}\frac{2}{k}=0<1

  • ⇒ Absolutely convergent (lecture example actually produced L=2L=2 ⇒ divergence; both demonstrate method — watch dominant powers).

  • Rule of thumb: any “something to the kk” strongly suggests root test.

Key Inequalities & Identities Used

  • Triangle inequality (finite sums): ∣∑<em>i=1Nc</em>i∣≤∑<em>i=1N∣c</em>i∣\Bigl|\sum<em>{i=1}^N c</em>i\Bigr|\le\sum<em>{i=1}^N |c</em>i|

  • Factorial definition: n!=n(n−1)(n−2)…1n! = n(n-1)(n-2)\dots1 with 0!=10! =1

  • Geometric-series sum (for ∣r∣<1|r|<1): ∑n=0∞arn=a1−r\sum_{n=0}^{\infty}ar^n=\dfrac{a}{1-r} (ratio tests mirror this behaviour).

Practical Strategy Checklist

  • Always start by checking lim⁡an\lim a_n; if ≠0 ⇒ done (diverges).

  • Look for obvious comparison to p-series or geometric series.

  • If term includes (–1)n(–1)^n ⇒ try Alternating-Series Test.

  • Factorials/exponentials ⇒ Ratio Test.

  • Powers like (expression)n(\text{expression})^{n} ⇒ Root Test.

  • When stuck, rewrite term: factor out largest nkn^k, simplify, or apply the continuous-function-commutes-with-limit trick.

  • Remember absolute vs conditional: test ∑∣an∣\sum|a_n| separately when signs alternate.

Recap: What Each Test Needs

  • Integral Test: f(x)≥0f(x)\ge0, decreasing, continuous; integrate f(x)f(x).

  • Comparison / Limit-Comparison: need a known benchmark series.

  • Divergence Test: just the limit of ana_n.

  • Alternating-Series: monotone ↓ and limit zero on positive part bnb_n.

  • Ratio / Root: compute limit of ratio or nth-root; compare with 1.

Prepare to mix & match — mastery comes from practising many examples (see added textbook scans & past exam). Happy studying!