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🎯⭐ INTERACTIVE LESSON

Infinite Sequences

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Infinite Sequences - Complete Interactive Lesson

Part 1: Definition & Convergence

Infinite Sequences

Part 1 of 7 — Definition & Convergence

A sequence is an ordered list of numbers: a1,a2,a3,…a_1, a_2, a_3, \ldots

Formally, a sequence is a function a:N→Ra: \mathbb{N} \to \mathbb{R} written as {an}\{a_n\}.

Convergence

lim⁡n→∞an=L  ⟹  the sequence converges to L\boxed{\lim_{n \to \infty} a_n = L \implies \text{the sequence converges to } L}

If no finite limit exists, the sequence diverges.

Examples

Sequence ana_nlim⁡n→∞an\lim_{n\to\infty} a_nConverges?
1/n1/n00Yes
(−1)n(-1)^nDNENo (oscillates)
n2n^2∞\inftyNo (unbounded)
(1+1/n)n(1 + 1/n)^neeYes
3+1/n23 + 1/n^233Yes

Key Fact: A sequence converges if and only if its terms approach a single finite number.

Computing Limits of Sequences

Technique 1: Direct substitution (polynomial/rational)

an=3n2+1n2−5→31=3a_n = \frac{3n^2 + 1}{n^2 - 5} \to \frac{3}{1} = 3 (divide by highest power of nn)

Technique 2: Squeeze Theorem

0≤sin⁡nn≤1n→00 \le \frac{\sin n}{n} \le \frac{1}{n} \to 0, so sin⁡nn→0\frac{\sin n}{n} \to 0.

Technique 3: L'Hôpital's via continuous extension

If f(x)f(x) is continuous and lim⁡x→∞f(x)=L\lim_{x \to \infty} f(x) = L, then lim⁡n→∞f(n)=L\lim_{n \to \infty} f(n) = L.

an=ln⁡nn:lim⁡x→∞ln⁡xx=L’Hlim⁡1/x1=0a_n = \frac{\ln n}{n}: \lim_{x \to \infty}\frac{\ln x}{x} \overset{\text{L'H}}{=} \lim \frac{1/x}{1} = 0

Technique 4: Root/ratio for exponential behavior

an=n!nn→0a_n = \frac{n!}{n^n} \to 0 (factorial grows slower than exponential of nn)

Practice Problems

Concept Checks

Computation

Summary

  • Sequence: ordered list {an}\{a_n\}
  • Converges if lim⁡n→∞an=L\lim_{n \to \infty} a_n = L (finite)
  • Techniques: direct comparison, squeeze theorem, L'Hôpital's, growth rate ordering
  • Growth rate hierarchy: ln⁡n≪np≪an≪n!≪nn\ln n \ll n^p \ll a^n \ll n! \ll n^n

Next: Part 2 — Monotone sequences and boundedness.

Part 2: Bounded & Monotonic Sequences

Infinite Sequences — Monotone & Bounded Sequences

Part 2 of 7 — Monotonicity, Bounds, and the Monotone Convergence Theorem

Definitions

PropertyMeaning
Increasingan+1≥ana_{n+1} \ge a_n for all nn
Decreasingan+1≤ana_{n+1} \le a_n for all nn
MonotoneEither increasing or decreasing
Bounded above∃M:an≤M\exists M: a_n \le M for all nn
Bounded below∃m:an≥m\exists m: a_n \ge m for all nn
BoundedBoth bounded above and below

Monotone Convergence Theorem

Monotone + Bounded  ⟹  Convergent\boxed{\text{Monotone + Bounded} \implies \text{Convergent}}

This is one of the most powerful tools for proving convergence without finding the limit.

Key Fact: An increasing sequence that is bounded above must converge. A decreasing sequence that is bounded below must converge.

Testing Monotonicity

Method 1: Difference test an+1−an>0  ⟹  a_{n+1} - a_n > 0 \implies increasing; <0  ⟹  < 0 \implies decreasing.

Method 2: Ratio test (for positive sequences) an+1/an>1  ⟹  a_{n+1}/a_n > 1 \implies increasing; <1  ⟹  < 1 \implies decreasing.

Method 3: Derivative test If an=f(n)a_n = f(n) and f′(x)>0f'(x) > 0 for x≥1x \ge 1, then {an}\{a_n\} is increasing.

Example

an=nn+1a_n = \frac{n}{n+1}. Is it monotone? Bounded?

an+1−an=n+1n+2−nn+1=(n+1)2−n(n+2)(n+2)(n+1)=1(n+2)(n+1)>0a_{n+1} - a_n = \frac{n+1}{n+2} - \frac{n}{n+1} = \frac{(n+1)^2 - n(n+2)}{(n+2)(n+1)} = \frac{1}{(n+2)(n+1)} > 0

So {an}\{a_n\} is increasing. Also an<1a_n < 1 for all nn (bounded above). By MCT, it converges. Indeed, lim⁡an=1\lim a_n = 1.

Practice Problems

Concept Checks

Computation

Summary

  • Monotone: always increasing or always decreasing
  • Test with: difference, ratio, or derivative
  • Monotone Convergence Theorem: Monotone + Bounded   ⟹  \implies Convergent
  • This theorem proves existence of a limit without finding it

Next: Part 3 — Recursive sequences and special limits.

Part 3: Geometric & Recursive Sequences

Infinite Sequences — Recursive Sequences & Special Limits

Part 3 of 7 — Recursion and Important Limits

Recursive Sequences

A recursive sequence defines an+1a_{n+1} in terms of previous terms:

an+1=f(an),a1=givena_{n+1} = f(a_n), \quad a_1 = \text{given}

To find the limit (if it converges), assume lim⁡an=L\lim a_n = L and solve:

L=f(L)L = f(L)

Example

a1=1a_1 = 1, an+1=2+ana_{n+1} = \sqrt{2 + a_n}.

If LL exists: L=2+L  ⟹  L2=2+L  ⟹  L2−L−2=0  ⟹  L=2L = \sqrt{2 + L} \implies L^2 = 2 + L \implies L^2 - L - 2 = 0 \implies L = 2 (since L>0L > 0).

Must also verify convergence: show the sequence is increasing and bounded above by 22.

Important Limits to Know

LimitValueWhy
lim⁡npan\lim \frac{n^p}{a^n} (a>1a > 1)00Exponential beats polynomial
lim⁡ann!\lim \frac{a^n}{n!}00Factorial beats exponential
lim⁡(ln⁡n)pnq\lim \frac{(\ln n)^p}{n^q} (q>0q > 0)00Polynomial beats logarithm
lim⁡n1/n\lim n^{1/n}11Apply ln⁡\ln: ln⁡nn→0\frac{\ln n}{n} \to 0
lim⁡n!nn\lim \frac{n!}{n^n}00nnn^n beats factorial
lim⁡(1+1/n)n\lim (1+1/n)^neeDefinition of ee
lim⁡rn\lim r^n ($r<1$)

Growth Rate Hierarchy

ln⁡n≪np≪an≪n!≪nn\boxed{\ln n \ll n^p \ll a^n \ll n! \ll n^n}

Each function on the left grows infinitely slower than the one on its right.

Practice Problems

Growth Hierarchy

Recursive Sequence

Summary

  • Recursive sequences: find limit by solving L=f(L)L = f(L)
  • Must separately verify convergence (monotone + bounded)
  • Growth hierarchy: ln⁡n≪np≪an≪n!≪nn\ln n \ll n^p \ll a^n \ll n! \ll n^n
  • Key limit: (1+k/n)n→ek(1 + k/n)^n \to e^k

Next: Part 4 — Sequences and series connection.

Part 4: Growth Rate Hierarchy

Infinite Sequences — Sequences vs. Series

Part 4 of 7 — The Bridge to Series

Sequence vs. Series

ConceptSymbolQuestion
Sequence{an}\{a_n\}Does an→La_n \to L?
Series∑an\sum a_nDoes ∑n=1∞an\sum_{n=1}^\infty a_n converge?

A series is the sum of a sequence. The partial sums form a new sequence:

SN=∑n=1Nan=a1+a2+⋯+aNS_N = \sum_{n=1}^{N} a_n = a_1 + a_2 + \cdots + a_N

∑n=1∞an=lim⁡N→∞SN\boxed{\sum_{n=1}^\infty a_n = \lim_{N \to \infty} S_N}

Key Fact: A series converges if and only if the sequence of partial sums converges.

The nnth Term Test (Divergence Test)

If lim⁡n→∞an≠0, then ∑an diverges.\boxed{\text{If } \lim_{n \to \infty} a_n \ne 0, \text{ then } \sum a_n \text{ diverges.}}

Contrapositive: If ∑an\sum a_n converges, then an→0a_n \to 0.

CAUTION: an→0a_n \to 0 does NOT guarantee convergence!

The harmonic series ∑1/n\sum 1/n has an=1/n→0a_n = 1/n \to 0 but diverges.

Geometric Series

∑n=0∞rn=11−rif ∣r∣<1\sum_{n=0}^\infty r^n = \frac{1}{1-r} \quad \text{if } |r| < 1

| ∣r∣|r| | Behavior | |-------|----------| | ∣r∣<1|r| < 1 | Converges to 11−r\frac{1}{1-r} | | ∣r∣≥1|r| \ge 1 | Diverges |

Practice Problems

Key Distinctions

Computation

Summary

  • Series = sum of a sequence: ∑an=lim⁡SN\sum a_n = \lim S_N
  • nnth Term Test: if an↛0a_n \not\to 0, series diverges
  • an→0a_n \to 0 does NOT guarantee convergence
  • Geometric series: converges iff ∣r∣<1|r| < 1, sum =a1−r= \frac{a}{1-r}

Next: Part 5 — Telescoping and pp-series.

Part 5: Sequences vs. Series

Infinite Sequences — Telescoping & pp-Series

Part 5 of 7 — Special Series Types

Telescoping Series

A telescoping series has partial sums where most terms cancel:

∑n=1∞(1n−1n+1)=lim⁡N→∞(1−1N+1)=1\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = \lim_{N\to\infty}\left(1 - \frac{1}{N+1}\right) = 1

How to recognize: Partial fractions often reveal telescoping structure.

pp-Series

∑n=1∞1np converges if and only if p>1\boxed{\sum_{n=1}^\infty \frac{1}{n^p} \text{ converges if and only if } p > 1}

SeriesppConverges?
∑1/n\sum 1/n11No (harmonic)
∑1/n2\sum 1/n^222Yes (=π2/6= \pi^2/6)
∑1/n\sum 1/\sqrt{n}1/21/2No
∑1/n3\sum 1/n^333Yes

AP Tip: The pp-series test and geometric series test are the most fundamental — many other tests compare to these.

Telescoping Example

Find ∑n=1∞1n(n+2)\sum_{n=1}^\infty \frac{1}{n(n+2)}.

Step 1. Partial fractions: 1n(n+2)=12(1n−1n+2)\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)

Step 2. Write partial sums: SN=12[(1−13)+(12−14)+(13−15)+⋯ ]S_N = \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots\right]

Most terms telescope! Surviving terms: SN=12(1+12−1N+1−1N+2)S_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right)

∑n=1∞1n(n+2)=12(32)=34\sum_{n=1}^\infty \frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{3}{2}\right) = \frac{3}{4}

Practice Problems

Classification

Computation

Summary

  • Telescoping series: write partial sums, identify cancellation
  • pp-series: ∑1/np\sum 1/n^p converges iff p>1p > 1
  • The harmonic series (p=1p = 1) is the critical boundary case
  • Partial fractions often reveal hidden telescoping

Next: Part 6 — Problem-Solving Workshop.

Part 6: Practice Workshop

Infinite Sequences — Workshop

Part 6 of 7 — Problem-Solving Workshop

Mixed problems covering sequence convergence, series basics, and special types.

Workshop Overview

Problem TypeKey Technique
Sequence limitGrowth hierarchy, L'Hôpital's
Recursive sequenceSolve L=f(L)L = f(L)
Telescoping sumPartial fractions, cancellation
Series classificationGeometric, pp-series, nnth term test

Problem Set

Problem 1. Does ∑n=1∞n2n2+1\sum_{n=1}^\infty \frac{n^2}{n^2 + 1} converge?

an=n2n2+1→1≠0a_n = \frac{n^2}{n^2+1} \to 1 \ne 0. Diverges by the nnth term test.

Problem 2. Find ∑n=0∞35n\sum_{n=0}^\infty \frac{3}{5^n}.

Geometric: a=3a = 3, r=1/5r = 1/5. Sum =31−1/5=34/5=154= \frac{3}{1-1/5} = \frac{3}{4/5} = \frac{15}{4}.

Problem 3. Classify ∑n=1∞1nπ\sum_{n=1}^\infty \frac{1}{n^\pi}.

pp-series with p=π≈3.14>1p = \pi \approx 3.14 > 1. Converges.

Workshop Questions

Quick Classification

Workshop Computation

Workshop Summary

  • nnth term test: quick divergence check (an↛0a_n \not\to 0)
  • Geometric and pp-series are the fundamental comparison targets
  • Growth hierarchy for sequence limits: ln⁡n≪np≪an≪n!\ln n \ll n^p \ll a^n \ll n!
  • Always check convergence before finding a sum

Next: Part 7 — Comprehensive Review.

Part 7: Final Assessment

Infinite Sequences — Comprehensive Review

Part 7 of 7 — Full Topic Review

Master Reference

TopicKey Result
Sequence convergencelim⁡an=L\lim a_n = L (finite)
Monotone ConvergenceMonotone + Bounded   ⟹  \implies Convergent
Recursive sequencesSolve L=f(L)L = f(L) for the limit
nnth Term Testan↛0  ⟹  ∑ana_n \not\to 0 \implies \sum a_n diverges
Geometric series∑arn=a1−r\sum ar^n = \frac{a}{1-r} if $
pp-series∑1/np\sum 1/n^p converges iff p>1p > 1
TelescopingUse partial fractions, find lim⁡SN\lim S_N
Growth hierarchyln⁡n≪np≪an≪n!≪nn\ln n \ll n^p \ll a^n \ll n! \ll n^n

AP Tip: The AP BC exam tests sequences primarily through series. Understanding sequence convergence is the foundation for all series work.

Common Pitfalls

  1. "an→0a_n \to 0 so ∑an\sum a_n converges" — FALSE. The harmonic series is the classic counterexample.

  2. Confusing the sequence {an}\{a_n\} with the series ∑an\sum a_n — one asks about the terms, the other about the sum.

  3. Forgetting to verify convergence of recursive sequences — solving L=f(L)L = f(L) only finds CANDIDATES for the limit.

  4. Incorrect geometric series formula — remember ∑n=0∞arn=a1−r\sum_{n=0}^\infty ar^n = \frac{a}{1-r} starts at n=0n = 0. If starting at n=1n = 1: ar1−r\frac{ar}{1-r}.

  5. pp-series boundary — p=1p = 1 (harmonic series) DIVERGES. Need p>1p > 1 (strictly).

Review Questions

Final Checks

Final Computation

Topic Complete!

You've mastered infinite sequences and the bridge to series:

  • Sequence convergence (limits, monotonicity, boundedness)
  • Recursive sequences and special limits
  • Geometric series, pp-series, and telescoping series
  • The nnth term test and its limitations

∑n=1∞an=lim⁡N→∞SN∑arn=a1−r  (∣r∣<1)\boxed{\sum_{n=1}^\infty a_n = \lim_{N \to \infty} S_N \qquad \sum ar^n = \frac{a}{1-r} \;(|r|<1)}

Up next: Infinite Series — convergence tests (comparison, integral, ratio, root).