Sequences¶
A sequence can be thought of as an ordered list of real numbers
$$
\big\{ a_1, a_2, a_3, ..., a_n, ... \big\}.
$$
In this notation $a_n$ is called the general term. When possible, the general term is given as an equation in terms of $n$. We will denote a sequence with general term $a_n$ by
$$
\Big\{a_n\Big\}_{n=1}^\infty \ \ \text{or}\ \ \big\{a_n\big\}.
$$
Note: It is not necessary that the first term in a sequence have index $1$. The index could start with any integer, but we usually request that it be non-negative. We will frequently choose to start sequences with $n=0$ as we move forward.
Moreover, regardless of the starting index that we choose, the general term of a sequence is generally not unique. This means that two students could come up with two different formulas for the general term that are not equivalent, but that both give the correct terms of the sequence for every value of $n$.
A sequence may also be regarded as a function whose domain is a subset of the natural numbers
$$
\mathbb{N} = \big\{0,1,2,3,...\big\},
$$
and whose codomain is the real numbers. The formula for the function is given by the general term,
$$
a(n) = a_n.
$$
1. Example¶
Find a formula for the general term of the sequence.
$$
\left\{ \frac{1}{2},-\frac{4}{3}, \frac{9}{4}, -\frac{16}{5}, \frac{25}{6}, ... \right\}
$$
Check your answer
One answer is
$$a_n = (-1)^{n+1}\,\frac{n^2}{n+1}$$
for $n = 1,2,3,...$
Video solution
Arithmetic and Geometric Sequences¶
Next, we consider two simple but important families of sequences.
An arithmetic sequence is a sequence for which adjacent terms have a common difference,
$$
a_{n+1} - a_n = d,
$$
for all $n$ in the domain of $a$. If the sequence is indexed starting at $n=0$, then arithmetic sequences have general term of the form
$$
a_n = a_0 + n\cdot d.
$$
Example. The sequence
$$\big\{ 7,11,15,19,23,27,... \big\}$$
has general term $a_n = 7 + 4n$.
A geometric sequence is a sequence for which adjacent terms have a common ratio,
$$
\frac{a_{n+1}}{a_n} = r,
$$
for all $n$ in the domain of $a$. If the sequence is indexed starting at $n=0$, then geometric sequences have general term of the form
$$
a_n = a_0 r^n.
$$
Example. The sequence
$$\left\{ 7,-1,\frac{1}{7},-\frac{1}{49},\frac{1}{343},... \right\}$$
has general term $a_n = 7\left(\frac{-1}{7}\right)^n$.
2. Example¶
Find arithmetic and geometric sequences, both starting at $n = 0$ with $a_0 = 5$ and $a_{10} = 15$.
Check your answer
The general term of the arithmetic sequence is $a_n = 5 + n$. The general term of the geometric sequence is $a_n = 5\cdot 3^{\frac{n}{10}}$.
Video solution
The Fibonacci Sequence¶
The general term is not always the simplest way to describe a sequence. Some sequences are best described by giving a relationship between their $n^{\text{th}}$ term and $(n+1)^{\text{st}}$ term. Such sequences are said to be defined recursively.
One famous example of a recursively defined sequence is the Fibonacci sequence, defined as follows.
The Fibonacci sequence is defined recursively by
$$
\begin{cases}
f_0 = 1,\\ f_1 = 1,\\
f_{n+2} = f_{n+1} + f_n,
\end{cases}
$$
for $n \geq 2$. The equation $f_{n+2} = f_{n+1} + f_n$ is called the recurrence relation of the sequence.
4. Exercise¶
Write out the first $10$ terms of the Fibonacci sequence.
Check your answer
$$\big\{ 1,1,2,3,5,8,13,21,34,55,... \big\}$$
Limits¶
Given a sequence $a :n \mapsto a_n$ it does not make sense to try to take a limit as $n$ approaches any finite value $n=k$ because the domain of the sequence is discrete. However, it does make sense to ask about the limit of the sequence as $n$ approaches $+\infty$.
Definition¶
A sequence $\left\{a_n\right\}$ has the limit $L$ and we write
$$
\lim a_n = \lim_{n\to \infty} a_n = L
$$
or simply $a_n \to L$ as $n \to \infty$, provided that for every every choice of $\varepsilon > 0$ there exists a corresponding integer $N = N(\varepsilon) > 0$ such that
$$
\big\vert a_n - L \big\vert < \varepsilon
$$
whenever $n > N$.
When a sequence has a limit, we say that the sequence converges. If a sequence does not have a limit (i.e., if its limit does not exist for any reason), then it it is said to diverge.
5. Exercise¶
Compare this definition with the $N-\varepsilon$ definition of the limit
$$
\lim_{x\to \infty} f(x)
$$
for a function $y = f(x)$ in Chapter 3 of the WebAssign eBook.
6. Example¶
Use the $N-\varepsilon$ definition of limit to prove that
$$
\lim \frac{3n}{1 + 6n} = \frac{1}{2}.
$$
Video solution
The Squeeze Theorem¶
The Squeeze Theorem of Calculus I applies to sequences in certain circumstances.
Theorem¶
Let $N_0 > 0$ be a fixed positive integer, $L$ a real number, and suppose $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ are sequences.
If $a_n \leq b_n \leq c_n$ for all $n > N_0$, and $\lim a_n = L = \lim c_n$, then
$$
\lim b_n = L.
$$
Corollary¶
Let $\{a_n\}$ be a sequence. If $\lim \vert a_n\vert = 0$, then $\lim a_n = 0$.
7. Exercise¶
Prove the corollary.
Solution
For all $n \geq 0$, $-\vert a_n\vert \leq a_n \leq \vert a_n\vert$. Since $\lim \vert a_n\vert$ is assumed to be $0$, then $\lim \big( - \vert a_n \vert \big) = 0$ as well. Therefore, by the Squeeze Theorem, $\lim a_n = 0$.
Exercises¶
Compute the limits
8. $\displaystyle \lim \frac{(-3)^n}{n!}$
Check your answer
$$0$$
9. $\displaystyle \lim 3^n7^{-n}$
Check your answer
$$0$$
10. $\displaystyle \lim \arctan(n)$
Check your answer
$$\frac{\pi}{2}$$
11. $\displaystyle \lim \left(1 + \frac{2}{n}\right)^n$
Check your answer
$$e^2$$
Video solution
12. $\displaystyle \lim \cos\left( \frac{n\pi}{n + 1} \right)$
Hint
Replace the squence with a continuous function on $\mathbb{R}$, then use continuity.
Check your answer
$$-1$$
Theorem¶
The sequence $\{r^n\}$ is convergent if and only if $-1 < r \leq 1$.
$$
\lim r^n = \begin{cases} 0 & \text{if}\ -1 < r < 1, \\
1 & \text{if}\ r = 1, \\
\text{DNE} & \text{otherwise}.
\end{cases}
$$
13. Exercise¶
Show that $\lim (-1)^n = \text{DNE}$.
Solution
If $r = -1$, the terms of the sequence $\big\{(-1)^n\big\}$ are
$$\Big\{1,-1,1,-1,1,-1,1,-1,1,-1, ... \Big\}.$$
Consider the limit of the sub-sequence of only the even indexed terms.
$$\lim\, (-1)^{2n} = \lim\, 1 = 1$$
On the other hand, consider the limit of the sub-seuqence of only the odd indexed terms.
$$\lim\, (-1)^{2n+1} = \lim\, (-1) = -1$$
Since the limits of these sub-sequences do not agree, thereore the limit does not exist.
Monotone Convergence Theorem¶
There will be many times when we wish to know whether or not a sequence is convergent, but we aren't necessarily concerned with the value that it converges to. In these situations there are tools available to us to help us answer our convergence questions. One of those tools is the Squeeze Theorem, mentioned in the notes above. A close cousin of the Squeeze Theorem is the Monotone Convergence Theorem.
Definition¶
A sequence $\{a_n\}$ is said to be bounded above if there is a number $M$ such that $a_n \leq M$ for all $n$ in the domain of $a$.
A sequence $\{a_n\}$ is said to be bounded below if there is a number $m$ such that $a_n \geq m$ for all $n$ in the domain of $a$.
A sequence $\{a_n\}$ is said to be bounded if and only if it is bounded both above and below,
$$
m \leq a_n \leq M
$$
for all $n$ in the domain of $a$.
Definition¶
A sequence $\{a_n\}$ is said to be increasing if $a_{n+1} > a_n$ for all $n$ in the domain of $a$.
A sequence $\{a_n\}$ is said to be decreasing if $a_{n+1} < a_n$ for all $n$ in the domain of $a$.
A sequence $\{a_n\}$ is said to be monotone (or monotonic) if and only if it is either increasing or decreasing.
Theorem¶
Every bounded, monotone sequence is convergent.
Video proof
This proof handles the case when $\{a_n\}$ is increasing and bounded above. You should sketch the proof of the case when $\{a_n\}$ is decreasing and bounded below as an exercise.
14. Example¶
Consider the sequence defined recursively by $a_0 = 2$ and $a_{n+1} = \tfrac{1}{2}\left(a_n + 6\right)$. Use the Monotone Convergence Theorem to show that this sequence converges, then compute the limit.
Video solution
Discussion¶
Questions? You can ask them below.