Series Convergence Tests — Lecture 3 Detailed Notes
Course Logistics
Practice material now online
1 authentic past mid-semester exam + 1 extra practice paper
Format, breadth & difficulty mirror upcoming Week-6 mid-semester exam
Weekly extra problems
Scans of textbook questions + selected solutions
Full solutions not posted → ask in office hours, drop-in centre, or demonstrator consultation
Assignment 1
Released end of Week 3 / start of Week 4
4–5 questions, each with multiple parts
Due sometime in Week 7 (after the mid-semester exam)
Lecture Focus
Few new theorems; mostly practice & technique
Complete leftover examples and add:
Alternating Series Test
Ratio Test
Root Test
Absolute vs Conditional Convergence
Integral Test (catch-up example)
Series: \sum_{k=1}^{\infty} k e^{-k}
Analogous function: f(x)=x e^{-x} (positive, continuous, decreasing for x\ge1)
Evaluate improper integral: \int_1^{\infty} x e^{-x}\,dx
Integration by parts
Let u=x\,,\; dv=e^{-x}dx\Rightarrow v=-e^{-x}
Result after limits: \frac{2}{e} (finite)
Integral converges ⇒ series converges by Integral Test
Key Technique Reminder
For rational functions inside limits: factor highest power of n in numerator & denominator leftover fractions \to0.
“Continuous Function commutes with limit” Trick
If f is continuous and \lim_{n\to\infty}a_n=L, then \lim_{n\to\infty}f(a_n)=f(L).
Example series: \sum_{n=1}^{\infty} \sqrt[3]{ \frac{n^2}{n^2+20n+9} }
Use cubic-root continuity: pull limit inside root
\lim_{n\to\infty} \frac{n^2}{n^2+20n+9}=1 ⇒ outer limit =1
Since term limit \neq0 ⇒ series diverges by Divergence Test (a.k.a. n\to\infty term test).
Choosing a Comparison (Example \sum \frac1{1+n^2})
Eyeball dominant term \sim\frac1{n^2} (p-series with p=2>1 ⇒ expect convergence).
Direct Comparison: \frac1{1+n^2} \le \frac1{n^2}.
Limit Comparison (slicker):
\lim_{n\to\infty}\frac{ \frac1{1+n^2} }{ \frac1{n^2} }=1 ⇒ same behaviour ⇒ convergent.Integral Test also works but more labour.
Alternating Series Test (Leibniz Criterion)
Series of form \sum_{n=1}^{\infty} (-1)^{n-1} b_n with
b_n>0
b_{n+1}\le b_n (monotonically decreasing)
\lim_{n\to\infty} b_n =0
⇒ Series converges (actually conditionally unless also absolutely conv.).
Classic Example: Alternating Harmonic Series
\sum_{n=1}^{\infty} (-1)^{n-1}\frac1n = \ln 2 (convergent although \sum 1/n diverges).
Counter-Examples / Quick Checks
If either monotonicity or b_n\to0 fails → cannot apply test; often diverges by \lim a_n test.
Example a_n=(-1)^n\frac{3n}{4n-1}: b_n\not\to0 ⇒ diverges.
Example (-1)^n \frac{n^2}{n^2+1}: monotone ↑ and \lim b_n=1 ⇒ diverges.
Absolute vs Conditional Convergence
Absolute Convergence: \sum |a_n| converges.
Conditional Convergence: Original series converges but \sum |a_n| diverges.
Relation: \text{Absolute} \Rightarrow \text{Convergent} by Triangle Inequality
\Big|\sum_{n=1}^{N} a_n\Big|\le\sum_{n=1}^{N}|a_n|.
Examples
(-1)^{n}\dfrac{3n}{4n-1} – not abs. (limit ≠0) and not cond. ⇒ diverges.
Alternating harmonic (-1)^{n-1}/n – convergent but not absolute ⇒ conditionally convergent.
(-1)^{n-1}/n^2 – p=2 so abs. conv. ⇒ both absolute & conditional.
Ratio Test
Define R =\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.
If R<1 ⇒ absolutely convergent.
If R>1 ⇒ divergent.
If R=1 ⇒ inconclusive.
Heuristic: compares with geometric series (common ratio R).
Good when series contains factorials, exponentials, large powers.
Example 1
a_n = (-1)^n\frac{n^3}{3^n}
\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^3}{3^{n+1}}\cdot\frac{3^{n}}{n^3}=\frac{(n+1)^3}{3 n^3}
Pull n^3: \to\frac13\Big(1+\frac1n\Big)^3 \xrightarrow[n\to\infty]{}\frac13<1 ⇒ absolutely convergent.
Example 2 (factorial)
a_n=\frac{n^n}{n!}
\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\frac{(n+1)^{n}}{n^n}\cdot\frac{1}{n+1}
Simplify: \Big(1+\frac1n\Big)^n\cdot\frac1{n+1}\to e\cdot0 = e>1
Actually final ratio =e>1 ⇒ diverges.
Root Test
Define L=\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}|a_n|^{1/n}.
L<1 ⇒ absolutely convergent.
L>1 ⇒ divergent.
L=1 ⇒ inconclusive.
Often simpler when term already has nth power.
Example
a_k=\left( \dfrac{2k-1}{k^2+3} \right)^{!k}, k\ge2
Root test: L=\lim_{k\to\infty}\left|a_k\right|^{1/k}=\lim_{k\to\infty}\frac{2k-1}{k^2+3}
Simplify: factor k/k^2 ⇒ L=\frac{2}{1}=2>1 ⇒ diverges.
How to Select a Test (Heuristics)
Divergence test first: if \lim a_n\neq0 ⇒ stop (diverges).
p-series / geometric resemblance ⇒ comparison or limit comparison.
Positive, monotone, integrable function ⇒ integral test.
Alternating signs ⇒ try Alternating Series Test.
Factorials, exponentials, n^n ⇒ Ratio Test.
a_n^n pattern ⇒ Root Test.
When absolute values easier ⇒ decide absolute vs conditional.
Ethical / Practical Notes
Copyright notice: material reproduced under Section 113P, further reproduction prohibited.
Use provided past exams & extra problems responsibly; seek help but do not share full solutions online.
Quick Formula & Reference Sheet
Triangle Inequality (finite sums): \Big|\sum_{k=1}^N x_k\Big|\le\sum_{k=1}^N |x_k|
Geometric series: \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\,(|r|<1)
p-series: \sum_{n=1}^{\infty} \frac1{n^p} converges \iff p>1.
Stirling (handy for factorial limits): n!\sim\sqrt{2\pi n}\,(n/e)^n (not used directly today but useful with Ratio/Root).
Big Take-aways
Master pattern recognition: decide test before heavy algebra.
Always check the easy tests (divergence, comparison) first.
Alternation aids convergence; absolute convergence is a stronger requirement.
Ratio & Root tests deliver fast verdicts when cancellation of large growth terms occurs.
Show all limit steps clearly; most proofs hinge on correct limit evaluation.