Chapter 5.3 Sequences

In this section, we develop the principle of finite approximation through sequences. We will define what a sequence is, how to describe sequences, and discuss properties certain sequences posses.

Analytical Properties of the Real Numbers

To begin, this is how we define a sequence.

Sequence

A sequence in a set \(X\) is a function \(a\) from \(\mathbb N\) to \(X\).

The standard convention is to denote by \((a_n)\) the sequence \(a\), where for all \(n\) in \(\mathbb N\), \[a(n) = a_n. \qquad \text{(read: "ay sub en" or "ay en")}\]

The term \(a_k\) is the \(k^{\rm th}\) term of the sequence, or the value of \(a\) in the \(k^{\rm th}\) place.

Use the figure below to understand what a sequence is.

Sequences arise in describing both terminating and non-terminating approximation procedures. First, understanding the meaning of the terms of a sequence with the example below.

Example 1

Write the first four terms of \((a_n)\), where for each natural number \(n\), \(a_n\) is given by:

  1. \(a_n = 2n-1\);

  2. \(a_n = n^2\);

  3. \(a_n = \tfrac{1}{3n}\)

  1. The first four terms:
    • \(a_1=2(1)-1=2-1=1\)
    • \(a_2=2(2)-1=4-1=3\)
    • \(a_3=2(3)-1=6-1=5\)
    • \(a_4=2(4)-1=8-1=7\)
  2. The first four terms:
    • \(a_1=(1)^2=1\)
    • \(a_2=(2)^2=4\)
    • \(a_3=(3)^2=9\)
    • \(a_4=(4)^2=16\)
  3. The first four terms:
    • \(a_1=\frac{1}{3\cdot 1}=\frac{1}{3}\)
    • \(a_2=\frac{1}{3\cdot 2}=\frac{1}{6}\)
    • \(a_3=\frac{1}{3\cdot 3}=\frac{1}{9}\)
    • \(a_4=\frac{1}{3\cdot 4}=\frac{1}{12}\)

A sequence does not have to be infinite. Here is what we mean by a finite sequence.

Finite Sequence

A finite sequence \(a\) in a set \(X\) is, for some natural number \(N\), a function from \(\{1, \dots, N\}\) to \(X\).

Such a sequence is denoted by \((a_n)_{n\in \{1, \dots, N\}}\).

It can be convenient to take a sequence to have domain equal to \(\mathbb N_0\) rather than \(\mathbb N\).

Recall that \[\mathbb N_0 = \{0, 1, 2, 3, \dots\}.\]

Understand this definition with the example below.

Example 2

The first six terms of \((a_n)_{n\in \{1, \dots, 6\}}\), where \[a_n = 3n-2,\] are \(a_1=1\), \(a_2=4\), \(a_3=7\) \(a_4=10\),\(a_5=13\) and \(a_6=16.\)

Just like a function, sequences can look and behave in diverse ways. In this next definition, we define what it means for a sequence to be “increasing” or “decreasing”.

Strictly Increasing, Increasing, Strictly Decreasing, Decreasing and Monotone

A sequence \((a_n)\) in \(\mathbb R\) is

  • strictly increasing if \(m>n\) implies that \(a_m > a_n\);

  • increasing if \(m>n\) implies that \(a_m \ge a_n\);

  • strictly decreasing if \(m>n\) implies that \(a_m< a_n\);

  • decreasing if \(m>n\) implies that \(a_m \leq a_n\).

A sequence \((a_n)\) in \(\mathbb R\) is (strictly) monotone if it is either (strictly) increasing or (strictly) decreasing.

Use the pictures below to understand the definition.

In the above statements of the definition, one can equivalently replace \(m\) with \(n+1\) and obtain the following equivalent definitions.

Equivalent Strictly Increasing and Strictly Decreasing Definition

If for every natural number \(n\), \[a_{n+1} > a_n,\] then \((a_n)\) is strictly increasing.

If for every natural number \(n\), \[a_{n+1} < a_n,\] then \((a_n)\) is strictly decreasing.

Use the next example to practice how to show a sequence is strictly increasing or strictly decreasing.

Example 3

Show that the sequence \((a_n)\) is increasing and the sequence \((b_n)\) is decreasing, where

  1. \(a_n = \tfrac{n}{n+1}\);

  2. \(b_n = \tfrac{n+2}{n+1}\)

  1. The \(n\) and \(n+1\) term are \(a_n=\tfrac{n}{n+1}\) and \(a_{n+1}=\tfrac{n+1}{n+2}\). To conclude \((a_n)\) is increasing, show for all \(n\) that \(a_{n+1}>a_n\). This is equivalent to showing that \(a_{n+1}-a_n>0\): \[ \begin{align*} a_{n+1}-a_n&=\tfrac{n+1}{n+2}-\tfrac{n}{n+1}\\ &=\tfrac{(n+1)(n+1)}{(n+2)(n+1)}-\tfrac{n(n+2)}{(n+2)(n+1)}\\ &=\tfrac{n^2+2n+1}{(n+2)(n+1)}-\tfrac{n^2+2n}{(n+2)(n+1)}\\ &=\tfrac{n^2+2n+1-n^2-2n}{(n+2)(n+1)}\\ &=\tfrac{1}{(n+2)(n+1)}. \end{align*} \] The expression \[a_{n+1}-a_n=\tfrac{1}{(n+2)(n+1)},\] is a quotient of two expressions that are always positive for all \(n\in\mathbb{N}\). Thus \(a_n\) is increasing.

  2. The \(n\) and \(n+1\) term are \(b_n=\tfrac{n+2}{n+1}\) and \(b_{n+1}=\tfrac{n+3}{n+2}\). To conclude \((b_n)\) is decreasing, show for all \(n\) that \(b_{n}>b_{n+1}\). This is equivalent to showing that \(b_{n}-b_{n+1}>0\):

\[ \begin{align*} b_{n}-b_{n+1}&=\tfrac{n+2}{n+1}-\tfrac{n+3}{n+2}\\ &=\tfrac{(n+2)(n+2)}{(n+2)(n+1)}-\tfrac{(n+1)(n+3)}{(n+2)(n+1)}\\ &=\tfrac{n^2+4n+4}{(n+2)(n+1)}-\tfrac{n^2+4n+3}{(n+2)(n+1)}\\ &=\tfrac{n^2+4n+4-n^2-4n-3}{(n+2)(n+1)}\\ &=\tfrac{1}{(n+2)(n+1)}. \end{align*} \] The expression \(b_{n}-b_{n+1}=\tfrac{1}{(n+2)(n+1)}\) is a quotient of two expressions that are always positive for all \(n\in\mathbb{N}\). Thus \(b_n\) is decreasing.

Any rational number can be explicitly expressed as a quotient of an integer by a natural number. Computing with rational numbers involves only arithmetic. However that is not true for all real numbers, specifically the irrational numbers.

Even though there are many more real numbers than rational numbers, largeness and smallness are still determined by natural numbers.

More precisely, \(\mathbb R\) has the Archimedean property.

Archimedan Property

Archimedan Property

There are two equivalent formulations.

  • For any real number \(x\), there is a natural number \(N\) so that \[N > x.\]

  • For any positive real number \(\varepsilon\), there is a natural number \(N\) so that \[0< \tfrac{1}{N} < \varepsilon.\]

Understand the definition with the following figure.

The first statement says for any real number \(x\), you can always find a big enough natural number \(N\) that is bigger than \(x\).

The second statement says that for any positive real number \(\varepsilon\), you can find a natural number \(N\) so its reciprocal is smaller than \(\varepsilon.\)

The Archimedean property of \(\mathbb R\) has some immediate consequences for infinite unions and intersections.

Archimedean Property Applied to Unions and Instructions

If \(x\) is a real number, then

  1. \(\displaystyle \bigcup_{n\in\mathbb N} \left(-\infty,x-\tfrac{1}{n}\right] = (-\infty, x)\);
  2. \(\displaystyle \bigcup_{n\in\mathbb N} \left[x+\tfrac{1}{n}, \infty\right) = (x, \infty)\);
  3. \(\displaystyle \bigcap_{n\in\mathbb N} \left(-\infty, x+\tfrac{1}{n}\right) = (-\infty, x]\);
  4. \(\displaystyle \bigcap_{n\in\mathbb N} \left(x-\tfrac{1}{n}, \infty\right) = [x, \infty)\).

Understand the consequences with the following picture.

Here are some examples of the consequences.

Example 4

We have the following

  1. Take \(I_n= \left(0-\tfrac{1}{n},\infty \right)\), then \(\displaystyle\bigcap_{n\in\mathbb N}I_n=[0,\infty)\);

  2. Take \(I_n= \left[2+\tfrac{1}{n},\infty \right)\), then \(\displaystyle\bigcup_{n\in\mathbb N}I_n=(2,\infty)\);

In the above example, we “computed” what the sequence of unions and sequence of intersections produced. This infinite process produced a tangible thing: in these examples an interval.

We give a name to this in the next subsection.

Sequential Limits and the Limit Laws

A sequence can be a collection of random numbers. It can also be a collection of numbers with some sort of pattern.

The terms of a sequence can get closer and closer to a single value, just like rational functions and exponential functions got closer and closer to their horizontal asymptotes.

We define this idea in the following definition.

Convergent Sequence

A sequence \((a_n)\) in \(\mathbb R\) converges to a real number \(L\) if:

  • For any positive real number \(\varepsilon\)
  • there is a natural number \(N\)
  • so that for any natural number \(n\),
  • if \(n\) is larger than \(N\), then \(|a_n - L| < \varepsilon\).

What does this rather complicated statement mean? It means that as long as \(n\) is large enough, \(a_n\) is as close as we like to \(L\).

Use the figure below to visualize this idea.

When a sequence converges to a real number, we use the following notation.

Limit of a Sequence

The number \(L\) is said to be the limit of the sequence \((a_n)\) and we denote this symbolically by \[\lim_{n\to \infty} a_n = L \quad{\rm or}\quad a_n \to L\] which reads: “the limit as \(n\) tends to infinity of \(a_n\) is equal to \(L\)” or “\((a_n)\) tends to \(L\) as \(n\) tends to \(\infty\).”

Here are two examples of a sequence.

Example 5

The sequences \(\big(\frac{1}{n}\big)\) and \(\big(-\frac{1}{n}\big)\) both converge to \(0\). Meaning, \[\lim_{n\to\infty}\frac{1}{n}=0\] and \[\lim_{n\to\infty}\left(-\frac{1}{n}\right)=0.\] The sequence \(a_n=\frac{1}{n}\) is decreasing toward 0 while \(b_n=-\frac{1}{n}\) increases toward 0.

Use these two plots to recognize why the limits are \(0\) for both.

A sequence does not have to converge to zero. In the next example, we construct two sequences that converge to non-zero numbers.

Example 6

Find an example of a sequence \((a_n)\) with the following properties:

  1. \((a_n)\) is strictly increasing and converges to 5;

  2. \((a_n)\) is strictly decreasing and converges to 7.

  1. The sequence \((a_n)\) defined by \(a_n=5-\frac{1}{n}\) is strictly increasing and converges to 5.

  1. The sequence \((a_n)\) defined by \(a_n=7+\frac{1}{n}\) is strictly decreases and converges to 7.

Here are two basic facts about the range of a convergent sequence in \(\mathbb R\).

Theorem on Convergent Sequences

  1. For any convergent sequence \((a_n)\), \(a(\mathbb N)\) is bounded. Meaning there are real numbers \(M\) and \(N\) so that \(N \leq a_n\leq M\) for all \(n\).

  2. For any sequence \((b_n)\) that converges to a nonzero real limit, if for all \(n\), \(b_n\) is nonzero, then there is positive real number \(\varepsilon\) so that \[|b_n| > \varepsilon.\]

Do you see why these statements are true?

They are true because if a sequence converges to a limit \(L\), then for any positive real number \(\varepsilon\), only finitely many terms of the sequence are outside the interval \((L-\varepsilon, L +\varepsilon)\), and so the limit “controls” all but finitely many terms of a convergent sequence.

To help us understanding more properties of a sequence, we first focus on a specific kind of sequence.

Null Sequence

The sequence \((a_n)\) is a null sequence if \((a_n)\) is convergent to \(0\).

It is convenient to prove statements about sequences by proving them first for null sequences. Think of null sequences as describing errors in finite approximation procedures. It is very useful to study such procedures by carefully studying the errors.

The equalities \[|a_n-0| = ||a_n|-0|,\quad |a_n-0| = |(a_n+L)-L| \quad \text{and} \quad |a_n-L| = |(a_n-L)-0|\] respectively imply the following:

Theorem on Null Sequences

  1. \((a_n)\) is null sequence if and only if \((|a_n|)\) is a null sequence;
  2. \((a_n)\) is a null sequence if and only if \((a_n +L)\) is convergent to \(L\);
  3. \((a_n)\) is convergent to \(L\) if and only if \((a_n -L)\) is a null sequence

The first statement tells us a sequence is null if and only if the absolute value of its terms is also a null sequence. The second statement tells us a sequence is null if and only if adding a constant to its terms converges to that constant. Finally, the third statement tell us that a sequence is convergent to \(L\) if and only if the difference of the sequence and its limit \(L\) is a null sequence.

We also have the following theorem as well.

Limit Laws for Null Sequences

  1. For any sequences \((a_n)\), \((b_n)\), and \((c_n)\) in \(\mathbb R\), if \((a_n)\) and \((c_n)\) are null sequences and for all \(n\), \[a_n \leq b_n \leq c_n,\] then \((b_n)\) is a null sequence.

  2. For any null sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\), \((a_n+b_n)\) is a null sequence.

  3. For any null sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\), \((a_nb_n)\) is a null sequence.

  4. For any sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\), if \((a_n)\) is a null sequence, if \((b_n)\) is convergent to a nonzero limit \(L\), and for all \(n\), if \(b_n\) is nonzero, then \(\big(\frac{a_n}{b_n}\big)\) is a null sequence.

Convergence theorems involving null sequences have a wide range of application, even though null sequences are very special kinds of sequences. As we will see, all convergence theorems for sequences that we will discuss follow from the convergence theorems that we have already presented.

The next examples shows the usefulness of explicitly writing error terms as a sequence.

Example 7

Take \(a_n\) to be the sequence that is given by \[a_n = 2^{\frac{1}{n}}.\] Calculate \(\displaystyle\lim_{n\to \infty} a_n\). Carefully justify your reasoning.

Based on some experiments, the limit should be \(1.\) This may be the case because the exponent of \(a_n\) is \(\frac{1}{n}\) and that approaches \(0\) as \(n\) approaches to infinity.

For each \(n \in\mathbb{N}\), there is a positive number \(\varepsilon_n\) (error term) so that \(2^{1/n}=1+\varepsilon_n\). In other words, \(2^{1/n}\) is just a little bit bigger than 1.

Use this to conclude that \[(2^{1/n})^n=(1+\varepsilon_n)^n\geq 1+n\varepsilon_n\] or \[2\geq n\varepsilon_n.\] Solve for \(\varepsilon\) to get \[0<\varepsilon_n<\frac{2}{n}.\] Since the error term is bounded by two null sequences, conclude by the Limit Laws for Null Sequences that \[\displaystyle\lim_{n\to\infty}\varepsilon_n=0.\] Thus, \[\displaystyle \lim_{n\to\infty}2^{1/n}=\displaystyle \lim_{n\to\infty}(1+\varepsilon_n)=1+0=1.\]

The next example also shows the usefulness of explicitly writing error terms as a sequence.

Example 8

Calculate \(\displaystyle\lim_{n\to \infty} \sqrt{9+\frac{1}{n}}\). Carefully justify your reasoning.

Based on some experiments, the limit should be \(3.\) This may be the case because the term \(\frac{1}{n}\) tends to 0, so the term \(9+\frac{1}{n}\) tends to \(9\).

For each \(n \in\mathbb{N}\), there is a positive number \(\varepsilon_n\) (error term) so that \(\sqrt{9+\frac{1}{n}}=3+\varepsilon_n\). This means that \[\left(\sqrt{9+\frac{1}{n}}\right)^2=(3+\varepsilon_n)^2\geq 3^2+2\cdot 3\varepsilon_n\] or \[9+\frac{1}{n}\geq 9+6\varepsilon_n.\] Solve for \(\varepsilon\) to get \[0<\varepsilon_n<\frac{1}{6n}.\] Since the error term is bounded by two null sequences, conclude by the Limit Laws for Null Sequences that \[\displaystyle\lim_{n\to\infty}\varepsilon_n=0.\] Thus, \[\displaystyle \lim_{n\to\infty}\sqrt{9+\frac{1}{n}}=\displaystyle \lim_{n\to\infty}(3+\varepsilon_n)=3+0=3.\]

General formulas, known as the limit laws, are very useful for calculating the limits of sequences.

Limit Law for Convergent Sequences

For any sequences \((a_n)\) and \((b_n)\) in \(\mathbb R\) that converge respectively to limits \(L\) and \(M\), \((a_n+b_n)\) and \((a_nb_n)\) are both convergent and

  1. (limit law for sums) \(\displaystyle\lim_{n\to\infty} (a_n+b_n) = L+M\);

  2. (limit law for products) \(\displaystyle\lim_{n\to\infty} a_nb_n=LM\).

  3. (limit law for quotients) In addition, if for all \(n\), \(b_n\) is nonzero and \(M\) is not equal to \(0\), then \(\left(\frac{a_n}{b_n}\right)\) is convergent and \(\displaystyle\lim_{n\to\infty} \frac{a_n}{b_n} = \frac{L}{M}\).

These results follow by rewriting the sequences like so.

  • \(a_n+b_n=a_n+b_n+L-L+M-M=(a_n-L)+(b_n-M)+L+M\),
  • \(a_nb_n =a_nb_n+Lb_n-b_nL+LM-LM=b_n(a_n-L)+L(b_n-M)+LM\)
  • \(\frac{a_n}{b_n}=\left(a_n\right)\cdot \left(\frac{1}{b_n}\right)=a_n\cdot \left(\frac{1}{b_n}-\frac{1}{M}+\frac{1}{M}\right)=a_n\cdot \left(\frac{1}{Mb_n}(b_n-M)+\frac{1}{M}\right)\)

We rewrite the above terms so that we introduce null sequences and can use the properties of null sequences in the previous theorem to obtain the results.

Example 9

Take \((a_n)\), \((b_n)\), and \((c_n)\) to be sequences with \[ \lim_{n\to\infty}a_n = 2, \quad \lim_{n\to\infty}b_n = 3, \quad {\rm and} \quad \lim_{n\to\infty}c_n = 7.\] Use the limit laws to compute the following:

  1. \(\displaystyle \lim_{n\to\infty}a_nb_nc_n\)

  2. \(\displaystyle \lim_{n\to\infty}a_n^3-5b_n\)

  3. \(\displaystyle \lim_{n\to\infty}\tfrac{a_n+b_n}{c_n}\)

  1. Use limit law for products to get \[\begin{align*} \lim_{n\to\infty}a_nb_nc_n&=\lim_{n\to\infty}\left(a_nb_n\right)c_n\\ &=\lim_{n\to\infty}\left(a_nb_n\right)\cdot \lim_{n\to\infty}c_n\\ &=\lim_{n\to\infty}a_n\cdot \lim_{n\to\infty} b_n\cdot \lim_{n\to\infty}c_n\\ &=2\cdot 3\cdot 7\\ &=42 \end{align*}\]

  2. Use limit law for products and sums to get \[\begin{align*} \lim_{n\to\infty}\left(a_n^3-5b_n\right)&=\lim_{n\to\infty}\left(a_n^3+(-5b_n)\right)\\ &=\lim_{n\to\infty}(a_n^3)+\lim_{n\to\infty}(-5\cdot b_n)\\ &=\lim_{n\to\infty}(a_n\cdot a_n\cdot a_n)+\lim_{n\to\infty}(-5\cdot b_n)\\ &=\lim_{n\to\infty}a_n\cdot \lim_{n\to\infty}a_n\cdot \lim_{n\to\infty}a_n+\lim_{n\to\infty}(-5)\cdot\lim_{n\to\infty} b_n\\ &=2\cdot 2\cdot 2+(-5)\cdot 3 \\ &=8-15\\ &=-7 \end{align*}\]

  3. Use limit law for quotient and sums to get \[\begin{align*} \lim_{n\to\infty}\tfrac{a_n+b_n}{c_n}&=\lim_{n\to\infty}\tfrac{(a_n+b_n)}{c_n}\\ &=\tfrac{\displaystyle\lim_{n\to\infty}(a_n+b_n)}{\displaystyle\lim_{n\to\infty}c_n}\\ &=\tfrac{\displaystyle\lim_{n\to\infty}a_n+\displaystyle\lim_{n\to\infty}b_n}{\displaystyle\lim_{n\to\infty}c_n}\\ &=\tfrac{2+3}{7}\\ &=\tfrac{5}{7} \end{align*}\]

In the next examples, we cannot use the limit laws immediately because the numerator and denominator do not converge. However, every sequence in the next example does converge. To see it, pay attention to how we rewrite the sequence.

Example 10

Use the limit laws to determine the following limits:

  1. \(\displaystyle \lim_{n\to\infty}\tfrac{5n+1}{n+3}\)
  2. \(\displaystyle \lim_{n\to\infty}\tfrac{n^2+7n+2}{n^2+2n-1}\)
  3. \(\displaystyle \lim_{n\to\infty}\tfrac{n-4}{n^2+3n-2}\)

  1. Rewrite the expression so the limit can be computed using limit laws. Multiply numerator and denominator by \(1=\tfrac{\frac{1}{n}}{\frac{1}{n}}\): \[\begin{align*} \frac{5n+1}{n+3}&=\frac{5n+1}{n+3}\cdot \tfrac{\frac{1}{n}}{\frac{1}{n}}\\ &=\frac{(5n+1)\cdot \frac{1}{n}}{(n+3)\cdot \frac{1}{n}}\\ &=\frac{5n\cdot \frac{1}{n}+1\cdot \frac{1}{n}}{n\cdot\frac{1}{n}+3\cdot\frac{1}{n}}\\ &=\frac{5+\frac{1}{n}}{1+3\cdot\frac{1}{n}}.\\ \end{align*}\] Now, use the limit laws: \[\begin{align*} \lim_{n\to\infty}\frac{5n+1}{n+3}&=\lim_{n\to\infty}\frac{5+\frac{1}{n}}{1+3\cdot\frac{1}{n}}\\ &=\frac{\displaystyle\lim_{n\to\infty}5+\displaystyle\lim_{n\to\infty}\frac{1}{n}}{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}3\cdot\displaystyle\lim_{n\to\infty}\frac{1}{n}}\\ &=\frac{5+0}{1+3\cdot0}\\ &=\frac{5}{1}\\ &=5. \end{align*}\]

  2. Rewrite the expression so the limit can be computed using limit laws. Multiply numerator and denominator by \(1=\tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\): \[\begin{align*} \frac{n^2+7n+2}{n^2+2n-1}&=\frac{n^2+7n+2}{n^2+2n-1}\cdot \tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\\ &=\frac{(n^2+7n+2)\cdot \frac{1}{n^2}}{(n^2+2n-1)\cdot \frac{1}{n^2}}\\ &=\frac{n^2\cdot \frac{1}{n^2}+7n\cdot \frac{1}{n^2}+2\cdot\frac{1}{n^2}}{n^2\cdot\frac{1}{n^2}+2n\cdot\frac{1}{n^2}-1\cdot\frac{1}{n^2}}\\ &=\frac{1+\frac{7}{n}+\frac{2}{n^2}}{1+\frac{2}{n}-\frac{1}{n^2}}\\ \end{align*}.\] Now, use the limit laws: \[\begin{align*} \lim_{n\to\infty}\tfrac{n^2+7n+2}{n^2+2n-1}&=\lim_{n\to\infty}\frac{1+\frac{7}{n}+\frac{2}{n^2}}{1+\frac{2}{n}-\frac{1}{n^2}}\\ &=\frac{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}\frac{7}{n}+\displaystyle\lim_{n\to\infty}\frac{2}{n^2}}{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}\frac{2}{n}-\displaystyle\lim_{n\to\infty}\frac{1}{n^2}}\\ &=\frac{1+0+0}{1+0-0}\\ &=\frac{1}{1}\\ &=1 \end{align*}.\]

  3. Rewrite the expression so the limit can be computed using limit laws. Multiply numerator and denominator by \(1=\tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\): \[\begin{align*} \frac{n-4}{n^2+3n-2}&=\frac{n-4}{n^2+3n-2}\cdot \tfrac{\frac{1}{n^2}}{\frac{1}{n^2}}\\ &=\frac{(n-4)\cdot \frac{1}{n^2}}{(n^2+3n-2)\cdot \frac{1}{n^2}}\\ &=\frac{n\cdot \frac{1}{n^2}-4\cdot \frac{1}{n^2}}{n^2\cdot\frac{1}{n^2}+3n\cdot\frac{1}{n^2}-2\cdot\frac{1}{n^2}}\\ &=\frac{\frac{1}{n}-\frac{4}{n^2}}{1+\frac{3}{n}-\frac{2}{n^2}}\\ \end{align*}.\] Now, use the limit laws: \[\begin{align*} \lim_{n\to\infty}\tfrac{n-4}{n^2+3n-2}&=\lim_{n\to\infty}\frac{\frac{1}{n}-\frac{4}{n^2}}{1+\frac{3}{n}-\frac{2}{n^2}}\\ &=\frac{\displaystyle\lim_{n\to\infty}\frac{1}{n}-\displaystyle\lim_{n\to\infty}\frac{4}{n^2}}{\displaystyle\lim_{n\to\infty}1+\displaystyle\lim_{n\to\infty}\frac{3}{n}-\displaystyle\lim_{n\to\infty}\frac{2}{n^2}}\\ &=\frac{0-0}{1+0-0}\\ &=\frac{0}{1}\\ &=0 \end{align*}.\]

In this example, we will show that if we have enough information, we can show a sequence converges. Pay close attention to how we use the limit laws. Notice that we do not use the limit law for products because we don’t know that \(b_n\) converges.

Example 11

Take \((a_n)\) and \((b_n)\) to be sequences with \[\lim_{n\to\infty}a_n=3\quad {\rm and} \quad \lim_{n\to\infty}a_nb_n=18.\] Carefully justify that \((b_n)\) is convergent and calculate its limit.

As long as \(a_n\not=0\), the identity \(b_n=\frac{a_nb_n}{a_n}\) holds. By Theorem on Convergent Sequences, there is an \(N\) so that \(a_n\not=0\) for \(n\geq N\). Use the limit laws to conclude that \[\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{a_nb_n}{a_n}=\frac{\displaystyle\lim_{n\to\infty}a_nb_n}{\displaystyle\lim_{n\to\infty}a_n}=\frac{18}{3}=6.\]

In this example, we will show that a sequences converges by rewriting it. Pay close attention to how and when we use the limit laws.

Example 12

Calculate \(\displaystyle\lim_{n\to \infty} n\left(\sqrt{9+\frac{1}{n}} - 3 \right).\) Carefully justify your reasoning.

The limit laws cannot be applied as written. Even though the conclusion of Example 9 was \(\displaystyle\lim_{n\to \infty}\left(\sqrt{9+\frac{1}{n}} - 3 \right)=3\), limit law for products cannot be applied because \(n\) does not converge. The limit laws for products only applies if both limits converge.

Instead, rewrite the expression in order to apply the limit laws. Notice that \(1=\frac{\sqrt{9+\frac{1}{n}} + 3}{\sqrt{9+\frac{1}{n}} + 3}.\) Use this to rewrite the expression:

\[ \begin{align*} n\left(\sqrt{9+\frac{1}{n}} - 3 \right)&=n\left(\sqrt{9+\frac{1}{n}} - 3 \right)\cdot\frac{\sqrt{9+\frac{1}{n}} + 3}{\sqrt{9+\frac{1}{n}} + 3}\\ &=n\cdot \frac{\left(\sqrt{9+\frac{1}{n}} - 3\right)\left(\sqrt{9+\frac{1}{n}} + 3\right)}{\sqrt{9+\frac{1}{n}}+3}\\ &=n\cdot \frac{9+\frac{1}{n}-9}{\sqrt{9+\frac{1}{n}}+3}\\ &=n\cdot \frac{\frac{1}{n}}{\sqrt{9+\frac{1}{n}}+3}\\ &= \frac{\frac{n}{n}}{\sqrt{9+\frac{1}{n}}+3}\\ &= \frac{1}{\sqrt{9+\frac{1}{n}}+3}.\\ \end{align*} \]

Now use the limit law of quotients to get

\[ \begin{align*} \displaystyle\lim_{n\to \infty} n\left(\sqrt{9+\frac{1}{n}} - 3 \right)&=\displaystyle\lim_{n\to \infty} \frac{1}{\sqrt{9+\frac{1}{n}}+3}\\ &=\frac{\displaystyle\lim_{n\to \infty}1}{\displaystyle\lim_{n\to \infty}\sqrt{9+\frac{1}{n}}+\displaystyle\lim_{n\to \infty}3}\\ &=\frac{1}{3+3}\\ &=\frac{1}{6}. \end{align*} \]

The squeeze theorem (often also called the sandwich theorem) is a valuable tool for determining the limit of a sequence.

To apply this theorem to determine the limit of a sequence \((b_n)\), the strategy is to find two sequences \((a_n)\) and \((c_n)\) so that for all \(n\), \[a_n \leq b_n\leq c_n\] and \((a_n)\) and \((c_n)\) converge to the same limit.

The sequence \((b_n)\) will then be convergent and will converge to the same limit as its bounding sequences.

Squeeze Theorem

The squeeze theorem states that if there are sequences \((a_n)\), \((b_n)\) and \((c_n)\) so that for all \(n\), \[a_n \leq b_n\leq c_n\] and \((a_n)\) and \((c_n)\) converge to the same limit, then \((b_n)\) will then be convergent and will converge to the same limit as its bounding sequences.

Finding such bounding sequences then becomes the main obstacle in applications. Use the squeeze theorem for all these problems.

Example 13

Take \(b_n\) to be the sequence that is given by \[b_n = (3^n +2^n)^{\frac{1}{n}}.\] Calculate \(\displaystyle\lim_{n\to \infty} b_n\). Carefully justify your reasoning.

Find two sequences \(a_n\) and \(c_n\) so that \(a_n\leq b_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}a_n=\displaystyle\lim_{n\to\infty}c_n.\) Based on the plot, the limit might be 3.

  • find upper bound: Notice that \(2^n\leq 3^n\) for all \(n\in\mathbb{N}\). So \[3^n+2^n\leq 3^n+3^n=2\cdot 3^n.\] Since the function \(f(x)=x^{1/n}\) is a monotic function when \(n\in\mathbb{N}\), \[3^n+2^n\leq 2\cdot 3^n\] implies that \[\begin{align*}\left(3^n+2^n\right)^{1/n}&\leq \left(2\cdot 3^{n}\right)^{1/n}\\&\leq 2^{1/n}\cdot3\end{align*}.\] Thus \(\left(3^n+2^n\right)^{1/n}\leq 3\cdot 2^{1/n}.\)

  • find lower bound: Now notice that \(2^n\) and \(3^n\) are always positive. So \[3^n+2^n\geq 3^n.\] Since the function \(f(x)=x^{1/n}\) is a monotic function when \(n\in\mathbb{N}\),\[3^n+2^n\geq 3^n\] implies that \[\begin{align*}\left(3^n+2^n\right)^{1/n}&\geq \left(3^n\right)^{1/n}\\&\geq 3\end{align*}.\] Thus \(\left(3^n+2^n\right)^{1/n}\geq 3.\)

Let \(a_n=3\) and \(c_n= 3\cdot 2^{n}\). The conclusion from the work before was \(a_n\leq b_n\leq c_n\) for all \(n\in\mathbb{N}\).

  • show upper and lower bound have same limit: \[\displaystyle \lim_{n\to\infty}a_n=\displaystyle \lim_{n\to\infty}3=3\] and \[\displaystyle \lim_{n\to\infty}c_n=\displaystyle \lim_{n\to\infty}\left(3\cdot 2^{1/n}\right)=\displaystyle \lim_{n\to\infty}3\cdot \displaystyle \lim_{n\to\infty}2^{1/n}=3\cdot 1=3.\]

Since \(a_n\leq b_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}a_n=\displaystyle\lim_{n\to\infty}c_n=3\), apply the squeeze theorem to conclude that \(\displaystyle\lim_{n\to\infty}(3^n +2^n)^{\frac{1}{n}}=3\).

In the next example, we show a sequence converges using the squeeze theorem. Pay attention to how we find the upper bounds and lower bounds.

Example 14

Take \(a_n\) to be the sequence that is given by \[a_n = \tfrac{4n^2 + n\sin(n)}{n^2+1}.\] Calculate \(\displaystyle\lim_{n\to \infty} a_n\). Carefully justify your reasoning.

Find two sequences \(b_n\) and \(c_n\) so that \(b_n\leq a_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}b_n=\displaystyle\lim_{n\to\infty}c_n.\) Based on the plot, the limit might be 4.

  • find upper bound: Recall that the sine function has a range of \([-1,1]\). This means that \(-1\leq \sin(n)\leq 1\) for all \(n\in\mathbb{N}\). Since \(4n^2\) and \(n\) are positive numbers, \[4n^2+n\sin(n)\leq 4n^2+n\cdot 1= 4n^2+n.\] Since \(n^2+1\) is always positive, \[4n^2+n\sin(n)\leq 4n^2+n\] implies that \[\frac{4n^2+n\sin(n)}{n^2+1}\leq \frac{4n^2+n}{n^2+1}.\]

  • find lower bound: Recall that the sine function has a range of \([-1,1]\). This means that \(-1\leq \sin(n)\leq 1\) for all \(n\in\mathbb{N}\). Since \(4n^2\) and \(n\) are positive numbers, \[4n^2+n\sin(n)\geq 4n^2+n\cdot(-1)= 4n^2-n.\] Since \(n^2+1\) is always positive, \[4n^2+n\sin(n)\geq 4n^2-n\] implies that \[\frac{4n^2+n\sin(n)}{n^2+1}\geq \frac{4n^2-n}{n^2+1}.\]

Let \(b_n=\frac{4n^2-n}{n^2+1}\) and \(c_n= \frac{4n^2+n}{n^2+1}\). The conclusion from the work before wast \(b_n\leq a_n\leq c_n\) for all \(n\in\mathbb{N}\).

  • show upper and lower bound have same limit: \(\displaystyle \lim_{n\to\infty}b_n=\displaystyle \lim_{n\to\infty}\frac{4n^2-n}{n^2+1}=4\) and \(\displaystyle \lim_{n\to\infty}c_n=\displaystyle \lim_{n\to\infty}\frac{4n^2-n}{n^2+1}=4\) (Work this out by following a similar idea in Example 10)

Since \(a_n\leq b_n\leq c_n\) and \(\displaystyle\lim_{n\to\infty}b_n=\displaystyle\lim_{n\to\infty}c_n=4\), apply the squeeze theorem to conclude that \(\displaystyle\lim_{n\to\infty}\tfrac{4n^2 + n\sin(n)}{n^2+1}=4\).

Not every sequence converges. In fact, we have already seen sequences that don’t converge. For example, \(a_n=n\).

We define how certain sequences behave when they don’t converge.

Diverges to Infinity

A sequence \((a_n)\) is said to diverge to \(\infty\) (read: “infinity”) if for any natural number \(M\), there is a natural number \(N\) so that for all \(n\) in \(\mathbb N\), \[n> N \quad \text{implies that} \quad a_n > M.\]

To indicate that \((a_n)\) diverges to \(\infty\), write \[\lim_{n\to \infty} a_n = \infty \quad \text{or}\quad a_n \to \infty;\]

This means that \(a_n\) is as large as we like as long as \(n\) is large enough.

A sequence can also diverge to negative infinity.

Diverges to Negative Infinity

A sequence \((a_n)\) is said to diverge to \(-\infty\) (read: “negative infinity”) if for any natural number \(M\), there is a natural number \(N\) so that for all \(n\) in \(\mathbb N\), \[n> N \quad \text{implies that} \quad a_n < -M.\]

To indicate that \((a_n)\) diverges to \(-\infty\), write \[\lim_{n\to \infty} a_n = -\infty \quad \text{or}\quad a_n \to -\infty.\]

Use the figure below to visualize this type of divergence.

People will often say that \((a_n)\) tends to \(\infty\) or that \((a_n)\) tends to \(-\infty\) to mean that it diverges to \(\infty\) or \(-\infty\), respectively.

In the next example, practice identifying what kind of divergence each sequence has.

Example 15

Determine whether the sequence diverges to either infinity or negative infinity:

  1. \(a_n = 2^n\);
  2. \(a_n = n^{1/3}\);
  3. \(a_n = -n^3\).
  4. \(a_n= n^2\)

  1. \(\displaystyle \lim_{n\to\infty}2^n=\infty\)

  1. \(\displaystyle \lim_{n\to\infty}n^{1/3}=\infty\)

  1. \(\displaystyle \lim_{n\to\infty}-n^3=-\infty\)

  1. \(\displaystyle \lim_{n\to\infty}n^2=\infty\)

Sequences may diverge, but not tend toward infinity or negative infinity

Example 16

Does the sequence \(a_n=\big((-1)^n\big)\) converge or diverge? If it converges, what is the limit. If it diverges, does it tend to infinity or negative infinity?

The sequence diverges. However, it does not diverge to infinity nor negative infinity.

Return

© Copyright 2024 by the POC Writing Team: Bryan Carrillo, Yat Sun Poon, and David Weisbart.

All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the POC Writing Team.