Sequences for MATH 141

Exam Relevance for MATH 141

Likelihood of appearing: High

Sequences open the series unit in MATH 141. Expect convergence problems using squeeze theorem and L'Hopital's.

Lesson

What is a Sequence?

A sequence is simply a list of numbers in a specific order. Unlike a set, order matters and repetition is allowed.

$$1, 4, 9, 16, 25, \ldots$$

This is the sequence of perfect squares. The "..." means the pattern continues forever.


Sequence Notation

There are several ways to write sequences:

1. List notation: $1, 4, 9, 16, 25, \ldots$

2. Set-builder notation: $\{a_n\}_{n=1}^{\infty}$ or simply $\{a_n\}$

3. Formula notation: $a_n = n^2$ for $n = 1, 2, 3, \ldots$

The Subscript $n$

The subscript $n$ is called the index. It tells you which term you're looking at:

  • $a_1$ = first term
  • $a_2$ = second term
  • $a_n$ = the $n$th term (general term)

Example: If $a_n = n^2$, then:

  • $a_1 = 1^2 = 1$
  • $a_2 = 2^2 = 4$
  • $a_3 = 3^2 = 9$
  • $a_{10} = 10^2 = 100$

Starting Index

Sequences don't always start at $n = 1$. Watch the notation!

$$\{a_n\}_{n=0}^{\infty} \text{ starts at } n = 0$$

$$\{a_n\}_{n=3}^{\infty} \text{ starts at } n = 3$$


Finding the Formula from a Pattern

Example 1: Find the General Term

Problem: Find a formula for $a_n$: $\quad 2, 4, 6, 8, 10, \ldots$

These are even numbers. The $n$th even number is $2n$.

$$\boxed{a_n = 2n}$$

Check: $a_1 = 2(1) = 2$ ✓, $a_2 = 2(2) = 4$ ✓, $a_5 = 2(5) = 10$ ✓


Example 2: Alternating Signs

Problem: Find a formula for $a_n$: $\quad 1, -1, 1, -1, 1, \ldots$

The signs alternate. Use $(-1)^n$ or $(-1)^{n+1}$:

  • $(-1)^n$ gives: $-1, 1, -1, 1, \ldots$ (starts negative)
  • $(-1)^{n+1}$ gives: $1, -1, 1, -1, \ldots$ (starts positive)

Since our sequence starts positive:

$$\boxed{a_n = (-1)^{n+1}}$$


Example 3: Alternating Signs with Another Pattern

Problem: Find a formula for $a_n$: $\quad -\frac{1}{2}, \frac{2}{3}, -\frac{3}{4}, \frac{4}{5}, \ldots$

Step 1: Identify the pattern without signs

Numerators: $1, 2, 3, 4, \ldots$ → $n$

Denominators: $2, 3, 4, 5, \ldots$ → $n + 1$

Step 2: Handle the alternating signs

First term is negative, so use $(-1)^n$:

$$\boxed{a_n = (-1)^n \cdot \frac{n}{n+1}}$$

Check: $a_1 = (-1)^1 \cdot \frac{1}{2} = -\frac{1}{2}$ ✓


Example 4: Factorials

Problem: Find a formula for $a_n$: $\quad 1, 2, 6, 24, 120, \ldots$

These are factorials:

  • $1 = 1!$
  • $2 = 2!$
  • $6 = 3!$
  • $24 = 4!$
  • $120 = 5!$

$$\boxed{a_n = n!}$$

Recall: $n! = n \cdot (n-1) \cdot (n-2) \cdots 3 \cdot 2 \cdot 1$


Limits of Sequences

The big question: What happens to $a_n$ as $n \to \infty$?

  • If $a_n$ approaches a finite number $L$, the sequence converges to $L$
  • If $a_n$ doesn't approach a finite number, the sequence diverges

$$\lim_{n \to \infty} a_n = L \quad \text{means the sequence converges to } L$$


Example 5: Convergent Sequence

Problem: Find $\displaystyle\lim_{n \to \infty} \frac{n}{n+1}$.

Divide numerator and denominator by $n$:

$$\lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}}$$

As $n \to \infty$, $\frac{1}{n} \to 0$:

$$= \frac{1}{1 + 0} = 1$$

$$\boxed{\lim_{n \to \infty} \frac{n}{n+1} = 1}$$

The sequence converges to 1.


Example 6: Divergent Sequence (Goes to Infinity)

Problem: Find $\displaystyle\lim_{n \to \infty} n^2$.

As $n$ increases, $n^2$ grows without bound:

$$\lim_{n \to \infty} n^2 = \infty$$

$$\boxed{\text{The sequence diverges (to } \infty\text{)}}$$


Example 7: Divergent Sequence (Oscillates)

Problem: Find $\displaystyle\lim_{n \to \infty} (-1)^n$.

The terms alternate: $-1, 1, -1, 1, -1, \ldots$

The sequence bounces between $-1$ and $1$ forever — it never settles on a single value.

$$\boxed{\text{The sequence diverges (oscillates)}}$$


Example 8: Sequence with Exponentials

Problem: Find $\displaystyle\lim_{n \to \infty} \frac{2^n}{n!}$.

This is a race between exponential growth ($2^n$) and factorial growth ($n!$).

Factorial wins! For large $n$, $n!$ grows much faster than $2^n$.

Write out some terms: $$\frac{2}{1}, \frac{4}{2}, \frac{8}{6}, \frac{16}{24}, \frac{32}{120}, \frac{64}{720}, \ldots$$

$$= 2, 2, 1.33, 0.67, 0.27, 0.089, \ldots$$

The terms are getting smaller and approaching 0.

$$\boxed{\lim_{n \to \infty} \frac{2^n}{n!} = 0}$$


Useful Limit Facts

For sequences, these limits are essential:

Sequence Limit
$\dfrac{1}{n^p}$ where $p > 0$ $0$
$r^n$ where $\|r\| < 1$ $0$
$r^n$ where $r > 1$ $\infty$
$r^n$ where $r \leq -1$ DNE (diverges)
$\dfrac{n^p}{a^n}$ where $a > 1$ $0$ (exponential beats polynomial)
$\dfrac{a^n}{n!}$ for any $a$ $0$ (factorial beats exponential)
$\sqrt[n]{n}$ $1$
$\left(1 + \dfrac{1}{n}\right)^n$ $e$

Example 9: Using Limit Facts

Problem: Find $\displaystyle\lim_{n \to \infty} \frac{n^{100}}{2^n}$.

This is polynomial ($n^{100}$) vs exponential ($2^n$).

Exponential always wins — no matter how large the polynomial power!

$$\boxed{\lim_{n \to \infty} \frac{n^{100}}{2^n} = 0}$$


Recursive Sequences

Some sequences are defined recursively — each term depends on previous terms.

Example: The Fibonacci sequence

$$a_1 = 1, \quad a_2 = 1, \quad a_n = a_{n-1} + a_{n-2} \text{ for } n \geq 3$$

This gives: $1, 1, 2, 3, 5, 8, 13, 21, \ldots$


Example 10: Computing Recursive Terms

Problem: Given $a_1 = 2$ and $a_{n+1} = 3a_n - 1$, find $a_4$.

$$a_1 = 2$$ $$a_2 = 3(2) - 1 = 5$$ $$a_3 = 3(5) - 1 = 14$$ $$a_4 = 3(14) - 1 = 41$$

$$\boxed{a_4 = 41}$$


Bounded and Monotonic Sequences

Bounded: A sequence is bounded if all terms stay within some fixed range.

  • Bounded above: There's a number $M$ where $a_n \leq M$ for all $n$
  • Bounded below: There's a number $m$ where $a_n \geq m$ for all $n$

Monotonic: A sequence is monotonic if it only goes one direction.

  • Increasing: $a_1 < a_2 < a_3 < \cdots$ (or $\leq$)
  • Decreasing: $a_1 > a_2 > a_3 > \cdots$ (or $\geq$)

Key Theorem: A bounded monotonic sequence always converges.


Example 11: Showing a Sequence is Increasing

Problem: Show that $a_n = \dfrac{n}{n+1}$ is increasing.

Method 1: Compare consecutive terms

$$a_{n+1} - a_n = \frac{n+1}{n+2} - \frac{n}{n+1}$$

$$= \frac{(n+1)^2 - n(n+2)}{(n+2)(n+1)}$$

$$= \frac{n^2 + 2n + 1 - n^2 - 2n}{(n+2)(n+1)} = \frac{1}{(n+2)(n+1)}$$

Since $\dfrac{1}{(n+2)(n+1)} > 0$ for all $n \geq 1$, we have $a_{n+1} > a_n$.

$$\boxed{\text{The sequence is increasing}}$$


Common Mistakes and Misunderstandings

❌ Mistake: Confusing $a_n$ with $a(n)$

Wrong: Treating $a_n$ as function notation $a(n)$ and writing $a_n = a \cdot n$.

Why it's wrong: The subscript $n$ is an index, not multiplication. $a_n$ means "the $n$th term of sequence $a$."

Correct: $a_n$ is read as "a sub n" — it's the term at position $n$.


❌ Mistake: Ignoring the starting index

Wrong: Assuming every sequence starts at $n = 1$.

Why it's wrong: $\{a_n\}_{n=0}^{\infty}$ has first term $a_0$, not $a_1$. The starting index changes which term is "first."

Correct: Always check the starting index. For $a_n = 2^n$ starting at $n = 0$: first term is $a_0 = 2^0 = 1$.


❌ Mistake: Saying a sequence "diverges to -1 and 1"

Wrong: "$(-1)^n$ diverges to both -1 and 1."

Why it's wrong: Divergence doesn't mean "going to multiple values." The sequence diverges because it fails to approach ANY single value.

Correct: "$(-1)^n$ diverges because it oscillates — it doesn't approach a limit."


❌ Mistake: Thinking polynomial can beat exponential for large n

Wrong: "$n^{1000}$ is huge, so $\frac{n^{1000}}{2^n} \to \infty$."

Why it's wrong: For large enough $n$, exponential growth ALWAYS overtakes polynomial growth. Always. No matter how large the exponent.

Correct: $\displaystyle\lim_{n \to \infty} \frac{n^{1000}}{2^n} = 0$. Exponential wins.

Formulas & Reference

Limit of a Sequence

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

If the terms a_n approach L as n gets larger and larger, the sequence converges to L. If no such L exists, the sequence diverges.

Variables:
$a_n$:
the nth term of the sequence
$L$:
the limit value (a finite number)
$n$:
the index (position in the sequence)

Limit of r^n

$$\lim_{n \to \infty} r^n = \begin{cases} 0 & \text{if } |r| < 1 \\ 1 & \text{if } r = 1 \\ \infty & \text{if } r > 1 \\ \text{DNE} & \text{if } r \leq -1 \end{cases}$$

The behavior of r^n depends on the value of r. This is fundamental for geometric sequences and series.

Variables:
$r$:
the common ratio
$|r|$:
absolute value of r

Definition of e (as a sequence limit)

$$\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$$

This sequence converges to Euler's number e ≈ 2.71828. Useful for compound interest and growth problems.

Variables:
$e$:
Euler's number ≈ 2.71828
$n$:
index approaching infinity

Factorial

$$n! = n \cdot (n-1) \cdot (n-2) \cdots 3 \cdot 2 \cdot 1$$

n factorial is the product of all positive integers from 1 to n. By convention, 0! = 1. Factorials grow faster than exponentials.

Variables:
$n!$:
n factorial
$0!$:
equals 1 by definition
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