MathLab with Mr. Ahmed

MathLab with Mr. Ahmed is a math learning website by Ahmed Amasha, a mathematics teacher teaching the American curriculum. Explore Algebra 2, Precalculus, Calculus, and SAT and PSAT Math Prep through lessons, focused practice and printable resources.

MathLabwith Mr. Ahmed
Infinite Series

9.1 Sequences

Focus on limits of sequences, monotone and bounded sequences, oscillating and divergent sequences.

Free preview Explore the lesson before creating an account.

Play games for this section

Before you solve

Ideas in this lesson

01Limits of sequencesWorked steps · five questions02Monotone and bounded sequencesWorked steps · five questions03Oscillating and divergent sequencesWorked steps · five questions
Idea 1 of 3

Limits of sequences

A sequence is evaluated at integer indices. Analyze its limiting terms without treating infinity as an ordinary index.

Worked example 1

Find the limit of an=2n+1n+2a_{n} = \dfrac{2n + 1}{n + 2} as n→∞n \to \infty .
  1. an=2+1n1+2na_{n} = \dfrac{2 + \dfrac{1}{n}}{1 + \dfrac{2}{n}}
  2. lim⁡n→∞an=2\displaystyle \lim\limits_{n\to \infty } a_{n} = 2

Answer: 22

Worked example 2

Find the limit of an=3n2a_{n} = \dfrac{3}{n^{2}} as n→∞n \to \infty .
  1. an=3n2a_{n} = \dfrac{3}{n^{2}}
  2. lim⁡n→∞an=0\displaystyle \lim\limits_{n\to \infty } a_{n} = 0

Answer: 00

Idea 2 of 3

Monotone and bounded sequences

A monotone bounded sequence converges. Increasing and decreasing refer to successive terms, not their signs.

Worked example 1

For n≥1,an=nn+2n \ge 1, a_{n} = \dfrac{n}{n + 2} is which type of sequence?
  1. an+1−an=2(n+2)(n+3)>0a_{n+1} - a_{n} = \dfrac{2}{(n + 2)(n + 3)} > 0
  2. 0<an<1  ⟹  0 < a_{n} < 1 \implies increasing and bounded

Answer: Increasing and bounded

Worked example 2

For n≥1,an=3nn \ge 1, a_{n} = \dfrac{3}{n} is which type of sequence?
  1. an+1−an=−3n(n+1)<0a_{n+1} - a_{n} = -\dfrac{3}{n(n + 1)} < 0
  2. 0<an≤3  ⟹  0 < a_{n} \le 3 \implies decreasing and bounded

Answer: Decreasing and bounded

Idea 3 of 3

Oscillating and divergent sequences

Two subsequences approaching different values prove divergence; boundedness alone is not sufficient.

Worked example 1

Does an=2(−1)na_{n} = 2(-1)^{n} converge?
  1. a2a_{2}ₖ =2= 2
  2. a2a_{2}ₖ+1=−2_{+1} = -2
  3. lim⁡a2\displaystyle \lim a_{2}ₖ ≠lim⁡a2\displaystyle \ne \lim a_{2}ₖ+1  ⟹  lim⁡\displaystyle _{+1} \implies \lim an_{n} does not exist

Answer: Diverges

Worked example 2

Does an=3(−1)na_{n} = 3(-1)^{n} converge?
  1. a2a_{2}ₖ =3= 3
  2. a2a_{2}ₖ+1=−3_{+1} = -3
  3. lim⁡a2\displaystyle \lim a_{2}ₖ ≠lim⁡a2\displaystyle \ne \lim a_{2}ₖ+1  ⟹  lim⁡\displaystyle _{+1} \implies \lim an_{n} does not exist

Answer: Diverges

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §9.1,p9.1, p. 584584. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.