Introduction to Sequences

Understanding sequences and their behavior

🎯 Introduction to Sequences

What is a Sequence?

A sequence is an ordered list of numbers:

a1,a2,a3,a4,,an,a_1, a_2, a_3, a_4, \ldots, a_n, \ldots

Each number is called a term.

Notation:

  • Individual terms: a1,a2,a3,a_1, a_2, a_3, \ldots
  • General term: ana_n (the nth term)
  • Whole sequence: {an}\{a_n\} or {an}n=1\{a_n\}_{n=1}^{\infty}

💡 Key Idea: A sequence is a function from positive integers to real numbers! an=f(n)a_n = f(n)


Explicit Formula

A formula that gives ana_n directly in terms of nn.

Example 1: an=1na_n = \frac{1}{n}

a1=1,a2=12,a3=13,a4=14,a_1 = 1, \quad a_2 = \frac{1}{2}, \quad a_3 = \frac{1}{3}, \quad a_4 = \frac{1}{4}, \ldots


Example 2: an=(1)nna_n = (-1)^n \cdot n

a1=1,a2=2,a3=3,a4=4,a_1 = -1, \quad a_2 = 2, \quad a_3 = -3, \quad a_4 = 4, \ldots

Alternating signs!


Recursive Formula

Defines each term using previous term(s).

Example 3: a1=1a_1 = 1, an=2an1a_n = 2a_{n-1} for n2n \geq 2

a1=1a_1 = 1 a2=2(1)=2a_2 = 2(1) = 2 a3=2(2)=4a_3 = 2(2) = 4 a4=2(4)=8a_4 = 2(4) = 8

This is the sequence 1,2,4,8,16,1, 2, 4, 8, 16, \ldots (powers of 2!)


Example 4 (Fibonacci): a1=1a_1 = 1, a2=1a_2 = 1, an=an1+an2a_n = a_{n-1} + a_{n-2} for n3n \geq 3

1,1,2,3,5,8,13,21,1, 1, 2, 3, 5, 8, 13, 21, \ldots


Finding Patterns

Given the first few terms, find the formula.

Example: 2,4,6,8,10,2, 4, 6, 8, 10, \ldots

Pattern: Even numbers!

an=2na_n = 2n


Example: 1,4,9,16,25,1, 4, 9, 16, 25, \ldots

Pattern: Perfect squares!

an=n2a_n = n^2


Example: 1,1,1,1,1,1, -1, 1, -1, 1, \ldots

Pattern: Alternating!

an=(1)n+1a_n = (-1)^{n+1} (or (1)n1(-1)^{n-1})


Limit of a Sequence

The sequence {an}\{a_n\} converges to limit LL if:

limnan=L\lim_{n \to \infty} a_n = L

Meaning: As nn gets larger, ana_n gets closer and closer to LL.


If the limit exists (finite): sequence converges

If the limit doesn't exist or is infinite: sequence diverges


Example 5: Find limn1n\lim_{n \to \infty} \frac{1}{n}

As nn \to \infty, the denominator grows without bound.

limn1n=0\lim_{n \to \infty} \frac{1}{n} = 0

The sequence converges to 0.


Example 6: Find limn2n+13n2\lim_{n \to \infty} \frac{2n + 1}{3n - 2}

Method: Divide numerator and denominator by highest power of nn (which is n1n^1).

limn2n+13n2=limn2nn+1n3nn2n\lim_{n \to \infty} \frac{2n + 1}{3n - 2} = \lim_{n \to \infty} \frac{\frac{2n}{n} + \frac{1}{n}}{\frac{3n}{n} - \frac{2}{n}}

=limn2+1n32n= \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{3 - \frac{2}{n}}

As nn \to \infty: 1n0\frac{1}{n} \to 0 and 2n0\frac{2}{n} \to 0

=2+030=23= \frac{2 + 0}{3 - 0} = \frac{2}{3}

The sequence converges to 23\frac{2}{3}.


Example 7: Find limn(1)n\lim_{n \to \infty} (-1)^n

The sequence is: 1,1,1,1,1,1,-1, 1, -1, 1, -1, 1, \ldots

It oscillates between -1 and 1, never settling down.

limn(1)n does not exist\lim_{n \to \infty} (-1)^n \text{ does not exist}

The sequence diverges.


Limit Laws for Sequences

If limnan=A\lim_{n \to \infty} a_n = A and limnbn=B\lim_{n \to \infty} b_n = B, then:

  1. Sum: limn(an+bn)=A+B\lim_{n \to \infty} (a_n + b_n) = A + B

  2. Difference: limn(anbn)=AB\lim_{n \to \infty} (a_n - b_n) = A - B

  3. Product: limn(anbn)=AB\lim_{n \to \infty} (a_n \cdot b_n) = A \cdot B

  4. Quotient: limnanbn=AB\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{A}{B} (if B0B \neq 0)

  5. Constant Multiple: limncan=cA\lim_{n \to \infty} c \cdot a_n = c \cdot A


Squeeze Theorem for Sequences

If anbncna_n \leq b_n \leq c_n for all nn, and:

limnan=limncn=L\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L

Then:

limnbn=L\lim_{n \to \infty} b_n = L


Example 8: Use Squeeze Theorem

Find limnsinnn\lim_{n \to \infty} \frac{\sin n}{n}.

Step 1: Recall 1sinn1-1 \leq \sin n \leq 1 for all nn.


Step 2: Divide by nn (positive):

1nsinnn1n-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}


Step 3: Take limits:

limn1n=0andlimn1n=0\lim_{n \to \infty} -\frac{1}{n} = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{1}{n} = 0


Step 4: By Squeeze Theorem:

limnsinnn=0\lim_{n \to \infty} \frac{\sin n}{n} = 0

The sequence converges to 0.


Monotonic Sequences

Increasing: anan+1a_n \leq a_{n+1} for all nn

Decreasing: anan+1a_n \geq a_{n+1} for all nn

Monotonic: either increasing or decreasing


Bounded Sequences

Bounded above: anMa_n \leq M for some MM and all nn

Bounded below: anma_n \geq m for some mm and all nn

Bounded: both bounded above and below


Monotone Convergence Theorem

If a sequence is monotonic and bounded, then it converges.

💡 Key Idea: If sequence is increasing but can't go past some ceiling, it must level off!


Example 9: Show an=nn+1a_n = \frac{n}{n+1} converges

Step 1: Find limit

limnnn+1=limn11+1n=11+0=1\lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = \frac{1}{1+0} = 1


Step 2: Check if increasing

an+1=n+1n+2a_{n+1} = \frac{n+1}{n+2}

Compare: Is n+1n+2nn+1\frac{n+1}{n+2} \geq \frac{n}{n+1}?

Cross-multiply: (n+1)2n(n+2)(n+1)^2 \geq n(n+2)

n2+2n+1n2+2nn^2 + 2n + 1 \geq n^2 + 2n

101 \geq 0

Sequence is increasing.


Step 3: Check if bounded

an=nn+1<1a_n = \frac{n}{n+1} < 1 for all nn

Also, an>0a_n > 0 for all nn.

Sequence is bounded: 0<an<10 < a_n < 1


Conclusion: By Monotone Convergence Theorem, sequence converges (to 1).


Common Sequence Limits

| Sequence | Limit | |----------|-------| | 1np\frac{1}{n^p} (p > 0) | 0 | | 1n\frac{1}{n} | 0 | | ana^n (if abs(a) < 1) | 0 | | ana^n (if a > 1) | ∞ | | lnnn\frac{\ln n}{n} | 0 | | n1/nn^{1/n} | 1 | | (1+1n)n\left(1 + \frac{1}{n}\right)^n | e |


L'Hôpital's Rule for Sequences

If limnan\lim_{n \to \infty} a_n has indeterminate form, treat nn as continuous variable xx and use L'Hôpital's Rule!

Example: limnlnnn\lim_{n \to \infty} \frac{\ln n}{n}

This is \frac{\infty}{\infty} form.

Replace nn with xx:

limxlnxx=limx1/x1=limx1x=0\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0


Example 10: limnn2en\lim_{n \to \infty} \frac{n^2}{e^n}

This is \frac{\infty}{\infty} form.

Use L'Hôpital's Rule (twice):

limxx2ex=limx2xex\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} (still \frac{\infty}{\infty})

=limx2ex=0= \lim_{x \to \infty} \frac{2}{e^x} = 0

The sequence converges to 0.

Exponentials grow faster than polynomials!


⚠️ Common Mistakes

Mistake 1: Confusing Sequence Index

a1a_1 is the FIRST term, not a0a_0 (unless stated otherwise).

Be careful with initial conditions!


Mistake 2: Thinking Convergence Means Limit is 0

WRONG: "If {an}\{a_n\} converges, then liman=0\lim a_n = 0"

EXAMPLE: an=1+1na_n = 1 + \frac{1}{n} converges to 1, not 0!


Mistake 3: Using Wrong Variable

When using L'Hôpital's Rule, replace discrete nn with continuous xx.


Mistake 4: Ignoring Oscillation

an=cos(nπ)a_n = \cos(n\pi) oscillates: 1,1,1,1,1, -1, 1, -1, \ldots

This diverges! Don't assume every formula converges.


📝 Practice Strategy

  1. Find first few terms to see the pattern
  2. For limits: divide by highest power, use L'Hôpital's if needed
  3. Check for oscillation: look for (1)n(-1)^n or trig functions
  4. Use Squeeze Theorem for bounded oscillating terms
  5. Monotonic + Bounded → converges (even if you can't find exact limit)
  6. Common limits: memorize 1np0\frac{1}{n^p} \to 0, rn0r^n \to 0 if r<1|r| < 1
  7. Exponentials beat polynomials: ene^n grows faster than nkn^k

📚 Practice Problems

1Problem 1medium

Question:

Find an explicit formula for the sequence 2,5,10,17,26,2, 5, 10, 17, 26, \ldots and determine if it converges.

💡 Show Solution

Step 1: Look for pattern in differences

Terms: 2, 5, 10, 17, 26

First differences: 3, 5, 7, 9

Second differences: 2, 2, 2

Since second differences are constant, this is a quadratic sequence!


Step 2: Assume form an=An2+Bn+Ca_n = An^2 + Bn + C

Use first three terms:

n=1n=1: A+B+C=2A + B + C = 2

n=2n=2: 4A+2B+C=54A + 2B + C = 5

n=3n=3: 9A+3B+C=109A + 3B + C = 10


Step 3: Solve system

From equations 1 and 2: (4A+2B+C)(A+B+C)=52(4A + 2B + C) - (A + B + C) = 5 - 2 3A+B=33A + B = 3 ... (equation i)

From equations 2 and 3: (9A+3B+C)(4A+2B+C)=105(9A + 3B + C) - (4A + 2B + C) = 10 - 5 5A+B=55A + B = 5 ... (equation ii)

Subtract (i) from (ii): 2A=22A = 2 A=1A = 1

From equation (i): B=33(1)=0B = 3 - 3(1) = 0

From original: C=2AB=210=1C = 2 - A - B = 2 - 1 - 0 = 1


Step 4: Formula is an=n2+1a_n = n^2 + 1

Check:

  • a1=1+1=2a_1 = 1 + 1 = 2
  • a2=4+1=5a_2 = 4 + 1 = 5
  • a3=9+1=10a_3 = 9 + 1 = 10

Step 5: Find limit

limn(n2+1)=\lim_{n \to \infty} (n^2 + 1) = \infty

The sequence diverges to infinity.


Answer: an=n2+1a_n = n^2 + 1, diverges

2Problem 2medium

Question:

Consider the sequence an=n212n2+3a_n = \frac{n^2 - 1}{2n^2 + 3}.

a) Find the first four terms. b) Determine if the sequence converges or diverges. If it converges, find the limit.

💡 Show Solution

Solution:

Part (a): Substitute n=1,2,3,4n = 1, 2, 3, 4:

a1=112+3=05=0a_1 = \frac{1 - 1}{2 + 3} = \frac{0}{5} = 0

a2=418+3=3110.273a_2 = \frac{4 - 1}{8 + 3} = \frac{3}{11} \approx 0.273

a3=9118+3=8210.381a_3 = \frac{9 - 1}{18 + 3} = \frac{8}{21} \approx 0.381

a4=16132+3=1535=370.429a_4 = \frac{16 - 1}{32 + 3} = \frac{15}{35} = \frac{3}{7} \approx 0.429

Part (b): Find limnan\lim_{n \to \infty} a_n:

Divide numerator and denominator by n2n^2:

limnn212n2+3=limn11n22+3n2\lim_{n \to \infty} \frac{n^2 - 1}{2n^2 + 3} = \lim_{n \to \infty} \frac{1 - \frac{1}{n^2}}{2 + \frac{3}{n^2}}

As nn \to \infty, terms with nn in denominator approach 0:

=102+0=12= \frac{1 - 0}{2 + 0} = \frac{1}{2}

The sequence converges to 12\frac{1}{2}.

3Problem 3easy

Question:

Determine if the sequence an=3n25n2n2+n+1a_n = \frac{3n^2 - 5n}{2n^2 + n + 1} converges. If so, find the limit.

💡 Show Solution

Step 1: Identify form

As nn \to \infty: both numerator and denominator → ∞

This is \frac{\infty}{\infty} form.


Step 2: Divide by highest power (which is n2n^2)

an=3n25n2n2+n+1=3n2n25nn22n2n2+nn2+1n2a_n = \frac{3n^2 - 5n}{2n^2 + n + 1} = \frac{\frac{3n^2}{n^2} - \frac{5n}{n^2}}{\frac{2n^2}{n^2} + \frac{n}{n^2} + \frac{1}{n^2}}

=35n2+1n+1n2= \frac{3 - \frac{5}{n}}{2 + \frac{1}{n} + \frac{1}{n^2}}


Step 3: Take limit as nn \to \infty

limnan=302+0+0=32\lim_{n \to \infty} a_n = \frac{3 - 0}{2 + 0 + 0} = \frac{3}{2}


Answer: The sequence converges to 32\frac{3}{2}.

4Problem 4hard

Question:

Prove that the sequence an=(1)nna_n = \frac{(-1)^n}{n} converges to 0.

💡 Show Solution

Step 1: Understand the sequence

a1=1,a2=12,a3=13,a4=14,a_1 = -1, \quad a_2 = \frac{1}{2}, \quad a_3 = -\frac{1}{3}, \quad a_4 = \frac{1}{4}, \ldots

Terms oscillate but get closer to 0.


Step 2: Use absolute value

an=(1)nn=1n|a_n| = \left|\frac{(-1)^n}{n}\right| = \frac{1}{n}


Step 3: Find limit of absolute value

limnan=limn1n=0\lim_{n \to \infty} |a_n| = \lim_{n \to \infty} \frac{1}{n} = 0


Step 4: Apply theorem

If limnan=0\lim_{n \to \infty} |a_n| = 0, then limnan=0\lim_{n \to \infty} a_n = 0.


Alternative: Use Squeeze Theorem

1n(1)nn1n-\frac{1}{n} \leq \frac{(-1)^n}{n} \leq \frac{1}{n}

limn1n=0andlimn1n=0\lim_{n \to \infty} -\frac{1}{n} = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{1}{n} = 0

By Squeeze Theorem:

limn(1)nn=0\lim_{n \to \infty} \frac{(-1)^n}{n} = 0


Answer: The sequence converges to 0 (proven by Squeeze Theorem or absolute value test).