Sequences for MATH 141
Exam Relevance for MATH 141
Sequences open the series unit in MATH 141. Expect convergence problems using squeeze theorem and L'Hopital's.
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
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: $a_1 = 2(1) = 2$ ✓, $a_2 = 2(2) = 4$ ✓, $a_5 = 2(5) = 10$ ✓
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:
$$\boxed{a_n = (-1)^{n+1}}$$
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: $a_1 = (-1)^1 \cdot \frac{1}{2} = -\frac{1}{2}$ ✓
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
$$\lim_{n \to \infty} a_n = L \quad \text{means the sequence converges to } L$$
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.
Problem: Find $\displaystyle\lim_{n \to \infty} n^2$.
As $n$ increases, $n^2$ grows without bound:
$$\lim_{n \to \infty} n^2 = \infty$$
$$\boxed{\text{The sequence diverges (to } \infty\text{)}}$$
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.
$$\boxed{\text{The sequence diverges (oscillates)}}$$
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) |
| $\sqrt[n]{n}$ | $1$ |
| $\left(1 + \dfrac{1}{n}\right)^n$ | $e$ |
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: $1, 1, 2, 3, 5, 8, 13, 21, \ldots$
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.
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.
Limit of a Sequence
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
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)
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 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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