## Section6.1Sequences

While the idea of a sequence of numbers, $a_1,a_2,a_3,\ldots$ is straightforward, it is useful to think of a sequence as a function. We have dealt with functions whose domains are the real numbers, or a subset of the real numbers, like $f(x)=\sin x\text{.}$ A sequence can be regarded as a function with domain as the natural numbers $\mathbb{N}=\{1,2,3,\ldots\}$ or the non-negative integers, $\ds \mathbb{Z}^{\geq 0}=\{0,1,2,3,\ldots\}\text{.}$ The range of the function is still allowed to be the set of all real numbers. We say that a sequence is a function $f\colon \mathbb{N}\to\R\text{.}$

###### Definition6.1. Infinite Sequence.

An infinite sequence of numbers is a function $f: \mathbb{N} \to \R$ such that

\begin{equation*} f(n) = a_n \end{equation*}

where

$a_1,\ a_2, \ a_3, \dots$ are the terms of the sequence, and

$a_n$ is the $n$-th term of the sequence.

Note: Sequences are commonly denoted in several different, but equally acceptable ways:
\begin{equation*} \begin{gathered} a_1,a_2,a_3,\ldots \\ \left\{a_n\right\}_{n=1}^\infty \\ \left\{f(n)\right\}_{n=1}^\infty \end{gathered} \end{equation*}

As with functions of the real numbers, we will most often encounter sequences that can be expressed by a formula. We have already seen the sequence $\ds a_i=f(i)=1-1/2^i\text{.}$ Some other simple examples are:

\begin{align*} f(i)\amp ={i\over i+1}\\ f(n)\amp ={1\over2^n}\\ f(n)\amp =\sin(n\pi/6)\\ f(i)\amp ={(i-1)(i+2)\over2^i} \end{align*}

Frequently these formulas will make sense if thought of either as functions with domain $\mathbb{R}$ or $\mathbb{N}\text{,}$ though occasionally one will make sense for integer values only.

The main question of interest when dealing with sequences is what happens to the terms as we go further and further down the list. In particular, as $i$ becomes extremely large, does $a_i$ get closer to one specific value? This is reminiscent of a question we asked in Chapter 3 of Calculus Early Transcendentals Differential & Multi-Variable Calculus for Social Sciences, when looking at limits of functions. In fact, the problems are closely related and we define the limit of a sequence in a way similar to the definition for Limit at Infinity in that chapter.

###### Definition6.2. Limit of a Sequence.

Suppose that $\ds\left\{a_n\right\}_{n=1}^\infty$ is an infinite sequence. We write

\begin{equation*} \lim_{n\to\infty}a_n = L\text{,} \end{equation*}

if $a_n$ can be made arbitrarily close to $L$ by taking $n$ large enough. If this limit exists, we say that the sequence converges to $L\text{,}$ otherwise it diverges.

Note: Intuitively, $\ds{\lim_{n\to\infty}a_n=L}$ means that the further we go in the sequence, the closer the terms get to $L\text{.}$

Interactive Demonstration. Investigate the three sequences $a_n$ below. Determine the limit $L=\ds\lim_{n\to\infty}a_n\text{,}$ if it exists.

###### Example6.3. Exponential Sequence.

Show that $\{2^{1/n}\}_{n=1}^{\infty}$ converges to 1.

Solution

We see that $a_n = 2^{1/n}$ and compute the following limit:

\begin{equation*} \lim_{n \to \infty} a_n = \lim_{n \to \infty} 2^{1/n} = 2^{0} = 1\text{.} \end{equation*}

Therefore, the sequence converges to 1.

If a sequence is defined by a formula $\{f(n)\}_{n=1}^{\infty}\text{,}$ we can often expand the domain of the function $f$ to the set of all (or almost all) real numbers. For example, $f(n)=\frac{1}{n}$ is defined for all non-zero real numbers.

When this happens, we can sometimes find the limit of the sequence $\{f(n)\}_{n=1}^{\infty}$ more easily by finding the limit of the function $f(x)\text{,}$ $x\in\mathbb{R}\text{,}$ as $x$ approaches infinity.

This follows immediately from the definition for Limit at Infinity in Chapter 3 of Calculus Early Transcendentals Differential & Multi-Variable Calculus for Social Sciences.

Note: Hereafter we will use the convention that $x$ refers to a real-valued variable and $i$ and $n$ are integer-valued.

###### Example6.5. Sequence of $1/n$.

Show that $\left\{\dfrac{1}{n}\right\}_{n=1}^{\infty}$ converges to 0.

Solution

Since $\ds\lim_{x\to\infty}\frac{1}{x}=0\text{,}$ $\ds\lim_{n\to\infty}\frac{1}{n}=0\text{.}$

Note:

1. The converse of Theorem 6.4 is not true. Let $f(n)=\sin(n\pi)\text{.}$ This is the sequence

\begin{equation*} \sin(0\pi), \sin(1\pi),\sin(2\pi),\sin(3\pi),\ldots=0,0,0,0,\ldots \end{equation*}

since $\sin(n\pi)=0$ when $n$ is an integer. Thus $\ds \lim_{n\to\infty}f(n)=0\text{.}$ But $\ds \lim_{x\to\infty}f(x)\text{,}$ when $x$ is real, does not exist: as $x$ gets bigger and bigger, the values $\sin(x\pi)$ do not get closer and closer to a single value, but take on all values between $-1$ and $1$ over and over.

2. In general, whenever you want to know $\ds \lim_{n\to\infty}f(n)$ you should first attempt to compute $\ds \lim_{x\to\infty}f(x)\text{,}$ since if the latter exists it is also equal to the first limit. But if for some reason $\ds \lim_{x\to\infty}f(x)$ does not exist, it may still be true that $\ds \lim_{n\to\infty}f(n)$ exists, but you'll have to figure out another way to compute it.

It is occasionally useful to think of the graph of a sequence. Since the function is defined only for integer values, the graph is just a sequence of points. In Figure 6.1 we see the graphs of two sequences and the graphs of the corresponding real functions.

Not surprisingly, the properties of limits of real functions translate into properties of sequences quite easily. The theorem on Limit Properties in Chapter 3 of Calculus Early Transcendentals Differential & Multi-Variable Calculus for Social Sciences becomes:

Likewise the Squeeze Theorem in Chapter 3 of Calculus Early Transcendentals Differential & Multi-Variable Calculus for Social Sciences becomes:

And a final useful fact:

This says simply that the size of $\ds |a_n|$ gets close to zero if and only if $\ds a_n$ gets close to zero.

###### Example6.9. Convergence of a Rational Function.

Determine whether $\ds\left\{{n\over n+1}\right\}_{n=0}^\infty$ converges or diverges. If it converges, compute the limit.

Solution

Defining $f(x)=\dfrac{x}{x+1}$ we obtain

\begin{equation*} \lim_{x\to\infty}{x\over x+1}=\lim_{x\to\infty}1-{1\over x+1}=1-0=1\text{.} \end{equation*}

Thus the sequence converges to 1.

###### Example6.10. Convergence of Ratio with Natural Logarithm.

Determine whether $\ds\bigg\{{\ln n\over n}\bigg\}_{n=1}^\infty$ converges or diverges. If it converges, compute the limit.

Solution

We define $f(x) = \dfrac{\ln x}{x}$ and compute

\begin{equation*} \lim_{x\to\infty}{\ln x\over x}=\lim_{x\to\infty}{1/x\over 1}= 0\text{,} \end{equation*}

using L'Hôpital's Rule. Thus the sequence converges to 0.

###### Example6.11. Alternating Terms.

Determine whether $\ds\{(-1)^n\}_{n=0}^\infty$ converges or diverges. If it converges, compute the limit.

Solution

$f(x)=(-1)^x$ is undefined for irrational values of $x$ so $\lim\limits_{x\to\infty}(-1)^x$ does not exist. However, the sequence has a very simple pattern:

\begin{equation*} 1,-1,1,-1,1\ldots \end{equation*}

and clearly diverges.

###### Example6.12. Convergence of Exponential Terms.

Determine whether $\ds\left\{\left(-1/2\right)^n\right\}_{n=0}^\infty$ converges or diverges. If it converges, compute the limit.

Solution

We consider the absolute value of the sequence $\ds\left\{\left|-1/2\right|^n\right\}_{n=0}^\infty=\left\{\left(1/2\right)^n\right\}_{n=0}^\infty$ and define

\begin{equation*} f(x) = \left(\frac{1}{2}\right)^x\text{.} \end{equation*}

Then

\begin{equation*} \lim_{x\to\infty}\left({1\over2}\right)^x=\lim_{x\to\infty}{1\over2^x}=0\text{,} \end{equation*}

so by the Absolute Value Sequence Theorem the sequence converges to 0.

###### Example6.13. Using the Squeeze Theorem for Sequences.

Determine whether $\ds\left\{\frac{\sin n}{\sqrt{n}}\right\}_{n=1}^\infty$ converges or diverges. If it converges, compute the limit.

Solution

Since $|\sin n|\le 1$ for all $n\text{,}$ we have that

\begin{equation*} 0\le\left|\frac{\sin n}{\sqrt{n}}\right|\le \frac{1}{\sqrt{n}}\text{.} \end{equation*}

Let $\ds a_n=0$ and $c_n=\dfrac{1}{\sqrt{n}}\text{,}$ then

\begin{equation*} \lim_{n\to\infty} a_n=\lim_{n\to\infty} 0=0 \text{ and } \lim_{n\to\infty}c_n = \lim_{n\to\infty} \frac{1}{\sqrt{n}}=0\text{.} \end{equation*}

Applying the Squeeze Theorem, we have

\begin{equation*} \lim_{n\to\infty} \frac{\sin n}{\sqrt{n}} = 0\text{,} \end{equation*}

and the sequence converges to 0.

###### Example6.14. Geometric Sequence.

Let $r$ be a fixed real number. Determine when $\{r^n\}_{n=0}^{\infty}$ coverges.

Solution

A particularly common and useful sequence is $\ds \{r^n\}_{n=0}^\infty\text{,}$ for various values of $r\text{.}$ Some are quite easy to understand:

• If $r=1$ the sequence converges to 1 since every term is 1, and likewise if $r=0$ the sequence converges to 0.

• If $r=-1$ this is the sequence of Example 6.11 and diverges.

• If $r>1$ or $r\lt -1$ the terms $\ds r^n$ get large without limit, so the sequence diverges.

• If $0\lt r\lt 1$ then the sequence converges to 0.

• If $-1\lt r\lt 0$ then $\ds |r^n|=|r|^n$ and $0\lt |r|\lt 1\text{,}$ so the sequence $\ds \{|r|^n\}_{n=0}^\infty$ converges to 0, so also $\ds\{r^n\}_{n=0}^\infty$ converges to 0.

In summary, $\ds \{r^n\}$ converges precisely when $-1\lt r\le1$ in which case

\begin{equation*} \lim_{n\to\infty} r^n=\left\{ \begin{array}{l l} 0 \amp \text{ if $-1\lt r\lt 1$ } \\ 1 \amp \text{ if $r=1$ } \end{array} \right. \end{equation*}

Sequences of this form, or the more general form $\{kr^n\}_{n=0}^{\infty}\text{,}$ are called geometric sequences or geometric progressions. They are encountered in a large variety of mathematical and real-world applications.

###### Definition6.15. Geometric Sequence.

A geometric sequence is of the form

\begin{equation*} \{a_n\}_{n=0}^{\infty} = \{kr^n\}_{n=0}^{\infty}\text{,} \end{equation*}

where $k$ is the first term and $r = \dfrac{a_{n+1}}{a_n}$ for $n \geq 0$ is the common ratio.

Sometimes we will not be able to determine the limit of a sequence, but we still would like to know whether it converges. In some cases we can determine this even without being able to compute the limit. For this, we introduce two basic properties of sequences, namely monotonicity and boundedness.

###### Definition6.16. Monotonic Sequence.

A sequence $\{a_n\}$ wih the following properties is called monotonic:

1. If $a_n \lt a_{n+1}$ for all $n\text{,}$ then the sequence is increasing or strictly increasing.

2. If $a_n \leq a_{n+1}$ for all $n\text{,}$ then the sequence is non-decreasing.

3. If $a_n > a_{n+1}$ for all $n\text{,}$ then the sequence is decreasing or strictly decreasing.

4. If $a_n \geq a_{n+1}$ then the sequence is non-increasing.

Unfortunately, some authors refer to a non-decreasing sequence as increasing or, similarly, to a non-increasing sequence as decreasing, which is not correct in the strict sense. For example, the sequence $\{a_1 = 0,\ a_2 = 0,\ a_3 = 1\}$ is not increasing (or strictly increasing), since $a_1=a_2=0\text{;}$ however, it is non-decreasing because $a_n \leq a_{n+1}$ for all $n\text{.}$
###### Example6.17. Monotonic or Not.

Determine the monotonicity of the following sequences:

1. $\left\{\dfrac{2^i-1}{2^i}\right\}_{i=1}^{\infty} = \dfrac{1}{2},\dfrac{3}{4}, \dfrac{7}{8},\dfrac{15}{16},\dots$

2. $\left\{\dfrac{n+1}{n}\right\}_{n=1}^{\infty} = \dfrac{2}{1}, \dfrac{3}{2},\dfrac{4}{3},\dfrac{5}{4},\dots$

Solution
1. We investigate monotonicity by comparing the $i$-th and $(i+1)$-th terms:

From the last statement, we observe that $-2 \lt -1\text{,}$ and so

\begin{equation*} \frac{2^i-1}{2^i} \lt \frac{2^{i+1}-1}{2^{i+1}} \end{equation*}

for all $i\text{.}$ Hence, the sequence is strictly increasing.

2. We need to compare the $n$-th and $n+1$-th terms:

From the last statement, we observe that $1 > 0$ and so

\begin{equation*} \frac{n+1}{n} > \frac{(n+1)+1}{(n+1)} \end{equation*}

for all $n\text{.}$ Hence, the sequence is strictly decreasing.

###### Definition6.18. Bounded Sequences.

A sequence $\{a_n\}$ is bounded above if there is some number $N$ such that $\ds a_n\le N$ for every $n\text{,}$ and bounded below if there is some number $N$ such that $\ds a_n\ge N$ for every $n\text{.}$ If a sequence is bounded above and bounded below it is bounded.

Note: If a sequence $\ds \{a_n\}_{n=0}^\infty$ is increasing or non-decreasing it is bounded below (by $\ds a_0$), and if it is decreasing or non-increasing it is bounded above (by $\ds a_0$).

Finally, with all this new terminology we can state an important theorem concerning the convergence of a monotonic and increasing sequence.

We will not prove this, but the proof appears in many calculus books. It is not hard to believe: suppose that a sequence is increasing and bounded, so each term is larger than the one before, yet never larger than some fixed value $N\text{.}$ The terms must then get closer and closer to some value between $\ds a_0$ and $N\text{.}$ It need not be $N\text{,}$ since $N$ may be a “too-generous” upper bound; the limit will be the smallest number that is above all of the terms $\ds a_i\text{.}$

###### Example6.20. Converging or Diverging.

Determine whether $\ds\left\{\frac{2^i-1}{2^i}\right\}_{i=1}^{\infty}$ converges.

Solution

For every $i\geq 1$ we have

\begin{equation*} 0\lt \frac{(2^i-1)}{2^i} \lt 1\text{,} \end{equation*}

so the sequence is bounded, and we have already observed that it is strictly increasing in Example 6.17. Therefore, the sequence converges according to the Bounded Monotonic Sequence Theorem.

Note: We don't actually need to know that a sequence is monotonic to apply this theorem — it is enough to know that the sequence is “eventually” monotonic, that is, that at some point it becomes increasing or decreasing. For example, the sequence $10\text{,}$ $9\text{,}$ $8\text{,}$ $15\text{,}$ $3\text{,}$ $21\text{,}$ $4\text{,}$ $3/4\text{,}$ $7/8\text{,}$ $15/16\text{,}$ $31/32,\ldots$ is not increasing, because among the first few terms it is not. But starting with the term $3/4$ it is increasing, so the theorem tells us that the sequence $3/4, 7/8, 15/16, 31/32,\ldots$ converges. Since convergence depends only on what happens as $n$ gets large, adding a few terms at the beginning can't turn a convergent sequence into a divergent one.

###### Example6.21. Demonstrating Convergence.

Show that $\ds\{n^{1/n}\}$ converges.

Solution

We first show that this sequence is decreasing by showing that

\begin{equation*} n^{1/n}> (n+1)^{1/(n+1)}\text{.} \end{equation*}

Consider the real function $\ds f(x)=x^{1/x}$ when $x\ge1\text{.}$ We can compute the derivative,

\begin{equation*} f'(x)=\frac{x^{1/x}(1-\ln x)}{x^2}\text{,} \end{equation*}

and note that when $x\ge 3$ this is negative. Since the function has negative slope,

\begin{equation*} \ds n^{1/n}> (n+1)^{1/(n+1)} \end{equation*}

when $n\ge 3\text{.}$

Since all terms of the sequence are positive, the sequence is decreasing and bounded when $n\ge3\text{,}$ and so the sequence converges. (As it happens, we can compute the limit in this case, but we know it converges even without knowing the limit; see Exercise 6.1.2.)

Note: When we need to determine whether a sequence $\{a_n\}$ is monotonic, it is often useful to investigate the ratio of successive terms $\dfrac{a_{n+1}}{a_n}\text{,}$ as shown in the following example.

###### Example6.22. Demonstrating Convergence.

Show that $\ds\left\{\frac{n!}{n^n}\right\}$ converges.

Solution

If we look at the ratio of successive terms we see that:

\begin{equation*} \begin{split} {a_{n+1}\over a_n} \amp= {(n+1)!\over (n+1)^{n+1}}{n^n\over n!}= {(n+1)!\over n!}{n^n\over (n+1)^{n+1}}\\ \amp= {n+1\over n+1}\left({n\over n+1}\right)^n= \left({n\over n+1}\right)^n \lt 1.\end{split} \end{equation*}

Therefore $a_{n+1}\lt a_n\text{,}$ and so the sequence is decreasing. Since all terms are positive, it is also bounded, and so it must converge. (Again it is possible to compute the limit; see Exercise 6.1.3.)

##### Exercises for Section 6.1.

Graph the first 5 terms of the following sequences.

1. $\ds\left\{n^2-n\right\}_{n=0}^{\infty}$

We plot $f(n) = n^2-n$ for $n=0,1,2,3,4$ below:
2. $\ds\left\{\frac{\sqrt{n}}{\sqrt{n}+1}\right\}_{n=0}^{\infty}$

We plot $f(n) = \frac{\sqrt{n}}{\sqrt{n}+1}$ for $n=0,1,2,3,4$ below:
3. $\ds\left\{\frac{e^n}{n^2}\right\}_{n=1}^{\infty}$

We plot $f(n)= \frac{e^n}{n^2}$ for $n=1,2,3,4,5$ below:
4. $\ds\left\{\frac{n}{3^n}\right\}_{n=1}^{\infty}$

We plot $f(n) = \frac{n}{3^n}$ for $n=1,2,3,4,5$ below:

Compute $\ds\lim_{x\to\infty} x^{1/x}\text{.}$Answer

$1$

Use the Squeeze Theorem to show that $\ds\lim_{n\to\infty} {n!\over n^n}=0\text{.}$

0
Solution
We first notice that
\begin{equation*} a_n = \frac{n!}{n^n} \geq \frac{1}{n^n}, \end{equation*}
and further that
\begin{equation*} a_n = \frac{n!}{n^2} = \frac{n(n-1)(n-2)\cdots 1}{n\cdot n \cdots n \cdot n} \leq \frac{1}{n}. \end{equation*}
Therefore, we have
\begin{equation*} \frac{1}{n^n} \leq a_n \leq \frac{1}{n}. \end{equation*}
Since both
\begin{equation*} \lim_{n\to \infty} \frac{1}{n^n} = 0 = \lim_{n\to\infty} \frac{1}{n}, \end{equation*}
by the Squeeze Theorem, it follows that
\begin{equation*} \lim_{n\to\infty} \frac{n!}{n^n} = 0. \end{equation*}

Determine whether the following sequences converge or diverge. In the case of convergence, compute the limit.

1. $\ds\{n^2-n\}_{n=0}^{\infty}$

diverges
Solution

$\lim\limits_{n\to\infty} \left(n^2-n\right) =\lim\limits_{n\to\infty} \left(n(n-1)\right)= \infty\text{,}$ and so the sequence $\{n^2-n\}_{n=0}^{\infty}$ diverges.

2. $\ds\left\{\frac{\sqrt{n}}{\sqrt{n}+1}\right\}_{n=0}^{\infty}$

converges to $1$
Solution

Let $f(x) = \dfrac{\sqrt{x}}{\sqrt{x}+1}\text{.}$ Then

\begin{equation*} \lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x}+1} = \lim_{x\to\infty}\frac{ \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} = 1\text{,} \end{equation*}

using L'Hôpital's rule. Therefore, the sequence $\left\{ \dfrac{\sqrt{n}}{\sqrt{n}+1}\right\}_{n=0}^{\infty}$ converges to 1.

3. $\ds\left\{\frac{e^n}{n^2}\right\}_{n=1}^{\infty}$

diverges
Solution

$\lim\limits_{n\to\infty} \dfrac{e^n}{n^2} = \infty\text{,}$ and so the sequence diverges.

4. $\ds\left\{\frac{n}{3^n}\right\}_{n=1}^{\infty}$

converges to $0$
Solution

Let $f(x) = \dfrac{x}{3^x}\text{.}$ Now by L'Hôpital's rule, we find

\begin{equation*} \lim_{x\to\infty} \frac{x}{3^x} = \lim_{x\to\infty} \frac{1}{3^x\log 3} = 0\text{.} \end{equation*}

The sequence $\left\{\dfrac{n}{3^n}\right\}_{n=0}^{\infty}$ thus converges to 0.

5. $\ds\left\{\sqrt{n+47}-\sqrt{n}\right\}_{n=0}^\infty$

converges to $0$
Solution
We rewrite:
\begin{equation*} a_n = \frac{\sqrt{n+47}-\sqrt{n}}{\sqrt{n+47}+\sqrt{n}} \left(\sqrt{n+47}+\sqrt{n}\right) = \frac{47}{\sqrt{n+47}+\sqrt{n}}. \end{equation*}
Therefore, since
\begin{equation*} \lim_{x\to\infty} \frac{47}{\sqrt{x+47}+\sqrt{x}} = 0, \end{equation*}
the sequence converges to 0.
6. $\ds\left\{{n^2+1\over (n+1)^2}\right\}_{n=0}^\infty$

converges to $1$
Solution
Consider
\begin{equation*} \lim_{x\to\infty} \frac{x^2+1}{(x+1)^2}. \end{equation*}
Then by applying L'Hôpital's rule twice, we find
\begin{equation*} \lim_{x\to\infty} \frac{x^2+1}{(x+1)^2} = \lim_{x\to\infty}\frac{2x}{2x+1} = \lim_{x\to\infty}{2}{2} = 1. \end{equation*}
Therefore, the sequence converges to 1.
7. $\ds\left\{{n+47\over\sqrt{n^2+3n}}\right\}_{n=1}^\infty$

converges to $1$
We notice that, for $n\geq 1\text{,}$
8. $\ds\left\{{2^n\over n!}\right\}_{n=0}^\infty$
converges to $0$
We bound $a_n$ by