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=1kek\sum_{k=1}^{\infty} k e^{-k}

  • Analogous function: f(x)=xexf(x)=x e^{-x} (positive, decreasing for x1x\ge1)

  • Evaluate improper integral 1xexdx\int_{1}^{\infty} x e^{-x}\,dx

    • Integration by parts

    • Let u=x,  dv=exdx    du=dx,  v=exu=x\,,\;dv=e^{-x}dx\;\Rightarrow\;du=dx\,,\;v=-e^{-x}

    • After computation:
      <em>1xexdx=[xexex]</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>na</em>n=L\lim<em>{n\to\infty} a</em>n=L, then lim<em>nf(a</em>n)=f(L)\lim<em>{n\to\infty} f(a</em>n)=f(L).

  • Example series: n=1n2n2+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:
      limnn2n2+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>na</em>n0\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=111+n2\sum_{n=1}^{\infty}\dfrac{1}{1+n^2}

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

  • Three possible strategies:

    1. Direct comparison: 11+n21n2\dfrac{1}{1+n^2}\le\dfrac{1}{n^2} ⇒ convergent.

    2. Limit Comparison (quickest):
      lim<em>n11+n21n2=lim</em>nn21+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)n1b</em>n\sum<em>{n=1}^{\infty}(-1)^{n-1} b</em>n with bn0b_n\ge0.

  • Converges if

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

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

  • Classic example: Alternating harmonic series
    n=1(1)n11n=ln2\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

limbn\lim b_n

Verdict

(1)n13n4n1\sum(-1)^{n-1}\,\dfrac{3n}{4n-1}

bnb_n decreasing

340\to\tfrac34\neq0

Divergent (fails limit)

(1)nn2n2+1\sum(-1)^{n}\,\dfrac{n^2}{n^2+1}

bnb_n increasing, 10\to1\neq0

Divergent

Alternating harmonic (1)n11n\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=1Na</em>n\Bigl|\sum<em>{n=1}^N a</em>n\Bigr|\le \sum<em>{n=1}^N |a</em>n|.

Examples

  • an=(1)n11na_n=(-1)^{n-1}\dfrac{1}{n}: conditional (series converges, absolute diverges).

  • an=(1)n11n2a_n=(-1)^{n-1}\dfrac{1}{n^2}: absolutely (and hence conditionally) convergent.

  • an=(1)n3n4n1a_n=(-1)^n\dfrac{3n}{4n-1}: neither (fails absolute; fails alternating-series test).


Ratio Test

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

    • If L<1 ⇒ absolutely convergent

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

Example 2 (factorials)

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

  • Ratio
    L=\lim{n\to\infty}\frac{(n+1)^{n+1}/(n+1)!}{n^n/n!}=\lim{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: an1/nnL\left|a_n\right|^{1/n}\xrightarrow[n\to\infty]{}L

    • L<1 ⇒ absolute convergence

    • L>1 ⇒ divergence

    • L=1L=1 ⇒ inconclusive

Example

k=2(2k1k2+3)!k\sum_{k=2}^{\infty}\left(\frac{2k-1}{k^2+3}\right)^{!k}

  • Apply root test:
    L=\lim{k\to\infty}\left|\frac{2k-1}{k^2+3}\right|=\lim{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=1Nc</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(n1)(n2)1n! = n(n-1)(n-2)\dots1 with 0!=10! =1

  • Geometric-series sum (for |r|<1): n=0arn=a1r\sum_{n=0}^{\infty}ar^n=\dfrac{a}{1-r} (ratio tests mirror this behaviour).


Practical Strategy Checklist

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

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

  • If term includes $(–1)^n$ ⇒ 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!