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Section7.1Sequences

Motivating Questions
  • What is a sequence?

  • What does it mean for a sequence to converge?

  • What does it mean for a sequence to diverge?

We encounter sequences every day. Your monthly utility payments, the annual interest you earn on investments, the amount you spend on groceries each week; all are examples of sequences. Other sequences with which you may be familiar include the Fibonacci sequence \(1, 1, 2, 3, 5, 8, \ldots \text{,}\) where each term is the sum of the two preceding terms, and the triangular numbers \(1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots \text{,}\) the number of vertices in the triangles shown on the right in Figure7.1.

Figure7.1Triangular numbers

Sequences of integers are of such interest to mathematicians and others that they have a journal1The Journal of Integer Sequences at http://www.cs.uwaterloo.ca/journals/JIS/ devoted to them and an online encyclopedia2The On-Line Encyclopedia of Integer Sequences at http://oeis.org/ that catalogs a huge number of integer sequences and their connections. Sequences are also used in digital recordings and images.

Our studies in calculus have dealt with continuous functions. Sequences model discrete instead of continuous information. We will study ways to represent and work with discrete information in this chapter as we investigate sequences and series, and ultimately see key connections between the discrete and continuous.

Example7.2

Suppose you receive \(\$5,\!000\) through an inheritance. You decide to invest this money into a fund that pays \(8\%\) annually, compounded monthly. That means that each month your investment earns \(\frac{0.08}{12} \cdot P\) additional dollars, where \(P\) is your principal balance at the start of the month. So in the first month your investment earns

\begin{equation*} 5000 \left(\frac{0.08}{12}\right) \end{equation*}

or \(\$ 33.33\text{.}\) If you reinvest this money, you will then have \(\$ 5033.33\) in your account at the end of the first month. From this point on, assume that you reinvest all of the interest you earn.

  1. How much interest will you earn in the second month? How much money will you have in your account at the end of the second month?

  2. Complete Table7.3 below to determine the interest earned and total amount of money in this investment each month for one year.

    Month Interest
    earned
    Total amount
    of money
    in the account
    \(0\) \(\$0.00\) \(\$5000.00\)
    \(1\) \(\$33.33\) \(\$5033.33\)
    \(2\)
    \(3\)
    \(4\)
    \(5\)
    \(6\)
    \(7\)
    \(8\)
    \(9\)
    \(10\)
    \(11\)
    \(12\)
    Table7.3Interest earned on an investment
  3. As we will see later, the amount of money \(P_n\) in the account after month \(n\) is given by

    \begin{equation*} P_n = 5000\left(1+\frac{0.08}{12}\right)^{n}\text{.} \end{equation*}

    Use this formula to check your calculations in Table7.3. Then find the amount of money in the account after 5 years.

  4. How many years will it be before the account has doubled in value to \(\$10,\!000\text{?}\)

Solution
  1. We will apply the monthly interest rate of \(\frac{.08}{12} \) to the balance at the end of month one. Thus, the interest earned during the second month will be

    \begin{equation*} \$ 5033.33\left(\frac{.08}{12}\right)=\$ 33.56. \end{equation*}

    This means that the account balance at the end of the second month will be \(\$ 5066.89.\)

  2. The completed table is as follows.

    Month Interest
    earned
    Total amount
    of money
    in the account
    \(0\) \(\$0.00\) \(\$5000.00\)
    \(1\) \(\$33.33\) \(\$5033.33\)
    \(2\) \(\$33.56\) \(\$5066.89\)
    \(3\) \(\$33.78\) \(\$5100.67\)
    \(4\) \(\$34.00\) \(\$5134.67\)
    \(5\) \(\$34.23\) \(\$5168.90\)
    \(6\) \(\$34.46\) \(\$5203.36\)
    \(7\) \(\$34.69\) \(\$5238.05\)
    \(8\) \(\$34.92\) \(\$5272.97\)
    \(9\) \(\$35.15\) \(\$5308.12\)
    \(10\) \(\$35.39\) \(\$5343.51\)
    \(11\) \(\$35.62\) \(\$5379.13\)
    \(12\) \(\$35.86\) \(\$5414.99\)
    Table7.4Interest earned on an investment
  3. You may notice some small difference in your results from the formula, due to the way rounding happens in each method. To compute the balance in the account after 5 years, we can use the formula with \(n=60\) months, to get

    \begin{equation*} P_{60}=5000\left(1+\frac{.08}{12}\right)^{60} = \$7449.23. \end{equation*}
  4. The goal is to find the time it takes for the account to reach a balance of \(\$ 10,\!000\text{.}\) To do this, we can use the formula, and filling in everything we know, we have

    \begin{equation*} 10000=5000\left(1+\frac{.08}{12}\right)^n. \end{equation*}

    We can then solve for the number of months as follows:

    \begin{align*} 10000 \amp = 5000\left(1+\frac{.08}{12}\right)^n\\ 2 \amp = \left(1+\frac{.08}{12}\right)^n\\ \ln (2) \amp = n \ln \left(1+\frac{.08}{12}\right)\\ \frac{\ln (2)}{\ln\left(1+\frac{.08}{12}\right)} \amp = n \text{.} \end{align*}

    This gives us that the number of months until the account reaches \(\$ 10,\!000\) is \(\frac{\ln (2)}{\ln\left(1+\frac{.08}{12}\right)} \approx 104.3 \text{.}\) This amounts to about \(8.7\) years.

SubsectionSequences

As Example7.2 illustrates, many discrete phenomena can be represented as lists of numbers (like the amount of money in an account over a period of months). We call any such list a sequence. A sequence is nothing more than list of terms in some order. We often list the entries of the sequence with subscripts,

\begin{equation*} s_1, s_2, \ldots, s_n, \ldots\text{,} \end{equation*}

where the subscript denotes the position of the entry in the sequence.

Sequence

A sequence is a list of terms \(s_1, s_2, s_3, \ldots\) in a specified order.

We can think of a sequence as a function \(f\) whose domain is the set of positive integers where \(f(n) = s_n\) for each positive integer \(n\text{.}\) This alternative view will be be useful in many situations.

We often denote the sequence \(s_1, s_2, s_3, \ldots\) by \(\{s_n\}\text{.}\) The value \(s_n\) (alternatively \(s(n)\)) is called the \(n\)th term in the sequence. If the terms are all 0 after some fixed value of \(n\text{,}\) we say the sequence is finite. Otherwise the sequence is infinite.

We often define sequences using formulas, which can be explicit or recursive. Explicit formulas write the term \(s_n\) in terms of \(n\text{,}\) while recursive formulas write \(s_n\) in terms of the earlier entries in the sequence. Often, explicit formulas are easier to work with because they can be used to directly find any term in the sequence.

Example7.5

Given the explicit sequence formula \(s_n=\frac{3n-1}{n+2}\text{,}\) find \(s_1, s_2, s_{10}\text{,}\) and \(s_{1000}\text{.}\)

Solution

Since the formula is explicit, we simply substitute in the appropriate value for \(n\text{.}\)

\begin{equation*} s_1= \frac{3(1)-1}{1+2}=\frac{2}{3}\text{,} \end{equation*}
\begin{equation*} s_2=\frac{3(2)-1}{2+2}=\frac{5}{4}\text{,} \end{equation*}
\begin{equation*} s_{10}=\frac{3(10)-1}{10+2}=\frac{29}{12}\text{,} \end{equation*}
\begin{equation*} s_{1000}=\frac{3(1000)-1}{1000+2}=\frac{2999}{1002}\text{.} \end{equation*}

Recursively defined sequences, on the other hand, use a formula that relies on previous terms in the sequence and requires some of the first few terms to be set to get started.

Example7.6

For each of the following recursively defined sequences, give the first six terms.

  1. \(s_n=s_{n-1}+2\) for \(n>1\) and \(s_1=2.\)

  2. \(s_n=-2s_{n-1}\) for \(n>1\) and \(s_1=1.\)

  3. \(s_n=s_{n-1}+2s_{n-2}\) for \(n>2\) and \(s_1=1, s_2=3\text{.}\)

Solution
  1. Notice that \(s_1=2\) was given and we'll need it to find \(s_2\text{.}\) For \(n=2\text{,}\)

    \begin{equation*} s_2=s_{2-1}+2=s_1+2=2+2=4. \end{equation*}

    Similarly, to get \(s_3\) we'll need to know \(s_2\text{.}\)

    \begin{align*} s_3\amp=s_2+2=4+2=6,\\ s_4\amp=s_3+2=6+2=8,\\ s_5\amp=s_4+2=8+2=10,\\ s_6\amp=s_5+2=12\text{.} \end{align*}
  2. This formula says that each term is the previous term multiplied by -2. We have

    \begin{align*} s_1\amp =1, \\ s_2\amp=-2(1)=-2, \\ s_3\amp=-2(-2)=4,\\ s_4\amp= -2(4)=-8, \\ s_5\amp=-2(-8)=16, \\ s_6\amp=-2(16)=-32 \text{.} \end{align*}
  3. Notice that this formula relies on two previous terms, not just one, which is why we had to give the first two terms ahead of time.

    \begin{align*} s_1\amp=1,\\ s_2\amp=3,\\ s_3\amp=s_2 + 2 s_1 = 3+2(1)=5,\\ s_4\amp=s_3+2s_2 = 5 + 2(3)=11,\\ s_5\amp=s_4+2s_3 = 11 + 2(5)=21,\\ s_6\amp=s_5+2s_4 = 21+2(11) = 43\text{.} \end{align*}

With infinite sequences, we are often interested in their end behavior and the idea of convergent sequences.

Example7.7
  1. Let \(s_n\) be the \(n\)th term in the sequence \(1, 2, 3, \ldots\text{.}\) Find a formula for \(s_n\) and use appropriate technological tools to draw a graph of entries in this sequence by plotting points of the form \((n,s_n)\) for some values of \(n\text{.}\) Most graphing calculators can plot sequences; directions follow for the TI-84.

    • In the MODEmenu, highlight SEQin the FUNCline and press ENTER.

    • In the Y=menu, you will now see lines to enter sequences. Enter a value for nMin (where the sequence starts), a function for u(n) (the \(n\)th term in the sequence), and the value of u_{n Min}.

    • Set your window coordinates (this involves choosing limits for \(n\) as well as the window coordinates XMin, XMax, YMin, and YMax.

    • The GRAPHkey will draw a plot of your sequence.

    Using your knowledge of limits of continuous functions as \(x \to \infty\text{,}\) decide if this sequence \(\{s_n\}\) has a limit as \(n \to \infty\text{.}\) Explain your reasoning.

  2. Let \(s_n\) be the \(n\)th term in the sequence \(1, \frac{1}{2}, \frac{1}{3}, \ldots\text{.}\) Find a formula for \(s_n\text{.}\) Draw a graph of some points in this sequence. Using your knowledge of limits of continuous functions as \(x \to \infty\text{,}\) decide if this sequence \(\{s_n\}\) has a limit as \(n \to \infty\text{.}\) Explain your reasoning.

  3. Let \(s_n\) be the \(n\)th term in the sequence \(2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \ldots\text{.}\) Find a formula for \(s_n\text{.}\) Using your knowledge of limits of continuous functions as \(x \to \infty\text{,}\) decide if this sequence \(\{s_n\}\) has a limit as \(n \to \infty\text{.}\) Explain your reasoning.

Hint
  1. When \(n=1\text{,}\) \(s_n = 1\text{;}\) when \(n=2\text{,}\) \(s_n = 2\text{.}\)

  2. Think about the value of \(\frac{1}{n}\text{.}\)

  3. Note that the numerator of each term of the sequence is one more than the denominator of the term.

Answer
  1. \(s_n = n\text{.}\) A plot of the first 20 points in the sequence is shown below.

    This sequence does not have a limit as \(n\) goes to infinity.

  2. \(s_n = \frac{1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

    This sequence has a limit of 0 as \(n\) goes to infinity.

  3. \(s_n = \frac{n+1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

    This sequence has a limit of 1 as \(n\) goes to infinity.

Solution
  1. By observation we see that a formula for \(s_n\) is \(s_n = n\text{.}\) A plot of the first 20 points in the sequence is shown below.

    We recall that a function \(f\) diverges to infinity (and hence does not have a limit3Limits when they exist are inherently finite. This is true even though we sometimes use the notation \(\lim\limits_{x\to\infty}f(x)=\infty\) to mean that a function increases without bound. as \(x\to\infty\)) if we can make the values of \(f(x)\) as large as we want by choosing large enough values of \(x\text{.}\) Since we can make the values of \(n\) in our sequence as large as we want by choosing \(n\) to be arbitrarily large, we suspect that this sequence does not have a limit as \(n\) goes to infinity. The fact that \(\lim\limits_{x\to\infty}x=\infty\) supports this reasoning.

  2. By observation we see that a formula for \(s_n\) is \(s_n = \frac{1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

    Since we can make the values of \(\frac{1}{n}\) in our sequence as close to 0 as we want by choosing \(n\) to be arbitrarily large, we suspect that this sequence has a limit of 0 as \(n\) goes to infinity. This is supported by the fact that \(\lim\limits_{x\to\infty}\frac1x=0\text{.}\)

  3. Since the numerator is always 1 more than the denominator, a formula for \(s_n\) is \(s_n = \frac{n+1}{n}\text{.}\) A plot of the first 20 points in the sequence is shown below.

    Since we can make the values of \(\frac{n+1}{n}\) in our sequence as close to 1 as we want by choosing \(n\) to be arbitrarily large, we suspect that this sequence has a limit of 1 as \(n\) goes to infinity. This is supported by the fact that \(\lim\limits_{x\to\infty}\frac{x+1}x=1\text{.}\)

Next we formalize the ideas from Example7.7.

Example7.8
  1. Recall our earlier work with limits involving infinity in Section3.6. State clearly what it means for a continuous function \(f\) to have a limit \(L\) as \(x \to \infty\text{.}\)

  2. Given that an infinite sequence of real numbers is a function from the integers to the real numbers, apply the idea from part (a) to explain what you think it means for a sequence \(\{s_n\}\) to have a limit as \(n \to \infty\text{.}\)

  3. Based on your response to part (b), decide if the sequence \(\left\{ \frac{1+n}{2+n}\right\}\) has a limit as \(n \to \infty\text{.}\) If so, what is the limit? If not, why not?

Hint
  1. A function \(f\) has limit \(L\) as \(x \to \infty\) if we can make \(f(x)\) as close to \(L\) as we like by \(\ldots\text{.}\)

  2. Think about making \(s_n\) as close as you want to \(L\) for large values of \(n\text{.}\)

  3. Consider the behavior of the function \(f(x) = \frac{1+x}{2+x}\) as \(x \to \infty\text{.}\)

Solution
  1. A continuous function \(f\) has a limit \(L\) as the independent variable \(x\) goes to infinity if we can make the values of \(f(x)\) as close to \(L\) as we want by choosing large enough values of \(x\text{.}\)

  2. We expect that a sequence \(\{s_n\}\) will have a limit \(L\) as \(n\) goes to infinity if we can make the entries \(s_n\) in the sequence as close to \(L\) as we want by choosing \(n\) to be sufficiently large.

  3. As \(n\) gets large, the constant terms become infinitesimally small compared to \(n\) and so \(\frac{1+n}{2+n}\) looks like \(\frac{n}{n}\) or 1 for large \(n\text{.}\) So the sequence \(\left\{ \frac{1+n}{2+n}\right\}\) has a limit of 1 at infinity.

    More rigorously, we can either write \(\frac{1+n}{2+n}=1-\frac1{2+n}\) and note that \(\lim\limits_{n\to\infty}\frac1{2+n}=0\text{,}\) or we can apply L'Hopital's rule to the function \(f(x)=\frac{1+x}{2+x}\) as \(x\to\infty\) to assert that \(\left\{\frac{1+n}{2+n}\right\}\) has limit \(1\) as \(n\to\infty\text{.}\)

In Example7.7 and Example7.8 we investigated a few sequences \(\{s_n\}\) that had a limit as \(n\) goes to infinity. More formally, we make the following definition.

Sequence Convergence

A sequence \(\{ s_n \}\) converges or is a convergent sequence provided that there is a number \(L\) such that we can make \(s_n\) as close to \(L\) as we want by taking \(n\) sufficiently large. In this situation, we call \(L\) the limit of the convergent sequence and write

\begin{equation*} \lim_{n \to \infty} s_n = L\text{.} \end{equation*}

If the sequence \(\{s_n\}\) does not converge, we say that the sequence \(\{s_n\}\) diverges.

The idea of sequence having a limit as \(n \to \infty\) is the same as the idea of a continuous function having a limit as \(x \to \infty\text{.}\) The only difference is that sequences are discrete instead of continuous.

We will apply the same terminology from limits of continuous functions to the limits of discrete functions. For example, we will say that a sequence converges to a limit \(L\) from below if the terms of the sequence are smaller than \(L\text{,}\) and from above if the terms of the sequence are larger than \(L\text{.}\)

Many properties of limits carry over from the continuous to discrete setting as well. These algebraic properties are essentially identical to the properties of limits listed in Section1.2:

Properties of Discrete Limits

Assuming all the limits on the right-hand side exist:

  • If \(b\) is a constant, then \(\lim\limits_{n \rightarrow \infty} (bs_n)=b\left(\lim\limits_{n \rightarrow \infty} s_n \right)\)
  • \(\lim\limits_{n \rightarrow \infty} \left( s_n+t_n\right)=\lim\limits_{n \rightarrow \infty} s_n+\lim\limits_{n \rightarrow \infty}t_n\)
  • \(\lim\limits_{n \rightarrow \infty} \left( s_n \cdot t_n\right)=\lim\limits_{n \rightarrow \infty} s_n\cdot\lim\limits_{n \rightarrow \infty}t_n\)
  • \(\lim\limits_{n \rightarrow \infty} \left(\frac{s_n}{t_n}\right)=\frac{\lim\limits_{n \rightarrow \infty} s_n}{\lim\limits_{n \rightarrow \infty} t_n}\text{,}\) provided \(\lim\limits_{n \rightarrow \infty} t_n \neq 0\)
  • For any constant \(k\text{,}\) \(\lim\limits_{n \rightarrow \infty} k=k\)

Example7.9

Use graphical and/or algebraic methods to determine whether each of the following sequences converges or diverges.

  1. \(\left\{\frac{1+2n}{3n-2}\right\}\)

  2. \(\left\{\frac{5+3^n}{10+2^n}\right\}\)

  3. \(\left\{\frac{10^n}{n!}\right\}\)4Note that \(!\) is the factorial symbol, and \(n! = n(n-1)(n-2) \cdots (2)(1)\) for any positive integer \(n\text{.}\) By convention, we define \(0!\) to be \(1\text{.}\)

Hint
  1. Multiply the numerator and denominator each by \(\frac{1}{n}\text{.}\)

  2. Compare the \(n\)th term to \(\frac{3^n}{2^n}\text{.}\)

  3. Plot the sequence and think about what the graph suggests.

Answer
  1. The sequence \(\left\{\frac{1+2n}{3n-2}\right\}\) converges to \(\frac{2}{3}\text{.}\)

  2. The sequence \(\left\{\frac{5+3^n}{10+2^n}\right\}\) diverges to infinity.

  3. \(\frac{10^n}{n!} \to 0\) as \(n \to \infty\text{.}\)

Solution
  1. A plot of the first 20 terms of the sequence \(\left\{\frac{1+2n}{3n-2}\right\}\) is shown below.

    The plot suggests that the sequence has a limit between 0.5 and 1. Evaluating the limit algebraically, we find that

    \begin{equation*} \lim\limits_{n\to\infty}\frac{1+2n}{3n-2}=\lim\limits_{n\to\infty}\frac{\frac1n+2}{3-\frac2n}=\frac23\text{.} \end{equation*}

    So the sequence \(\left\{\frac{1+2n}{3n-2}\right\}\) converges to \(\frac{2}{3}\text{.}\)

  2. A plot of the first 20 terms of the sequence \(\left\{\frac{5+3^n}{10+2^n}\right\}\) is shown below. Note the scale on the vertical axis.

    The plot implies that the sequence does not have a limit as \(n\) goes to infinity. Evaluating the limit algebraically, we factor \(3^n\) from the numerator and \(2^n\) from the denominator. Doing so yields

    \begin{equation*} \lim\limits_{n\to\infty}\frac{5+3^n}{10+2^n}=\lim\limits_{n\to\infty}\frac{3^n\left(\frac{5}{3^n}+1\right)}{2^n\left(\frac{10}{2^n}+1\right)}\text{.} \end{equation*}

    Now, since \(\lim\limits_{n\to\infty}\frac{\frac5{3^n}+1}{\frac{10}{2^n}+1}=1\text{,}\) it follows that

    \begin{equation*} \lim\limits_{n\to\infty}\frac{3^n\left(\frac{5}{3^n}+1\right)}{2^n\left(\frac{10}{2^n}+1\right)}=\lim\limits_{n\to\infty}\left(\frac32\right)^n\text{.} \end{equation*}

    Since \(\frac{3}{2} \gt 1\text{,}\) this limit is infinite and the sequence \(\left\{\frac{5+3^n}{10+2^n}\right\}\) diverges to infinity as the plot suggests.

  3. A plot of the first 20 terms of the sequence \(\left\{\frac{10^n}{n!}\right\}\) is shown below. Note the scale on the vertical axis.

    Initially, it looks as though the terms increase without bound, but beginning at about \(n=10\) the factorial in the denominator dominates the numerator. Notice that

    \begin{equation*} \frac{10^n}{n!} = \frac{10 \times 10 \times 10 \times \cdots \times 10}{1 \times 2 \times 3 \times \cdots \times n}\text{.} \end{equation*}

    When \(n \gt 20\text{,}\) we have that \(\frac{10}{n} \lt \frac{1}{2}\) and thus

    \begin{align*} \frac{10^n}{n!} \amp = \left(\frac{10 \times 10 \times 10 \times \cdots \times 10}{1 \times 2 \times 3 \times \cdots\times 20}\right) \left(\frac{10 \times 10 \times 10 \times \cdots \times 10}{21 \times 22 \times 23 \times \cdots\times n}\right)\\ \amp = \left(\frac{10^{20}}{20!}\right) \left(\frac{10}{21}\right) \left(\frac{10}{22}\right) \cdots \left(\frac{10}{n}\right)\\ \amp \lt \left(\frac{10^{20}}{20!}\right) \left(\frac{1}{2}\right)^{n-20}\text{.} \end{align*}

    Since \(\frac{1}{2}\lt 1\text{,}\) the term \(\left(\frac{1}{2}\right)^{n-20}\) goes to 0 as \(n\) goes to infinity. The fact that \(\frac{10^{20}}{20!}\) is a constant means that \(\frac{10^n}{n!} \to 0\) as \(n \to \infty\text{.}\)

There are some special types of sequences for which convergence is sometimes easier to determine.

Bounded Sequence

A sequence \(\{s_n\}\) is bounded if there are numbers \(U\) and \(L\) such that \(U \leq s_n \leq L \) for all \(n\text{.}\) The number \(U\) is called a upper bound, while \(L\) is called a lower bound.

Monotone Sequence

A sequence is monotone if it is either increasing or decreasing. That is, if either \(s_n \leq s_{n+1} \) for all \(n\text{,}\) or \(s_n \geq s_{n+1}\) for all \(n\text{.}\)

Example7.10

Determine whether the following sequences are bounded, monotone, both, or neither. Then, determine whether the sequence converges or diverges.

  1. \(s_n = \left(\frac{1}{3}\right)^n \)

  2. \(s_n = (-1)^n \)

  3. \(s_n = 2^n +1 \)

  4. \(s_n= \frac{1}{2}s_{n-1}\) for \(n>1\) and \(s_1=30\text{.}\)

  5. \(s_n = (-2)^n \)

  6. \(s_n=\frac{(-1)^n}{n} \)

  7. \(s_n = \left( 1+\frac{1}{n}\right) \)

Solution
  1. This sequence is bounded because \(0 \leq \left(\frac{1}{3}\right)^n \leq \frac{1}{3} \) for all \(n\geq 1\text{.}\) It's also decreasing, so it's monotone as well. It converges and has limit 0.

  2. This sequence is bounded because \(-1 \leq (-1)^n \leq 1 \) for all \(n\text{.}\) However, since \(s_n\) oscillates between \(-1\) and \(1\text{,}\) it is not monotone. The sequence diverges because of the oscillation.

  3. \(2^n +1\) is unbounded because it gets arbitrarily large as \(n\) gets large. It is monotone increasing. The sequence diverges because it's unbounded.

  4. This recursive sequence is bounded between \(0\) and \(30\text{,}\) and since each term is half the previous one, it is monotone decreasing. The sequence converges to \(0\text{.}\)

  5. \((-2)^n \) is neither bounded nor monotone. It is not bounded below or above, and is not monotone since it alternates between positive and negative terms. The sequence diverges.

  6. This sequence is not monotone, since it alternates between positive and negative terms. It is bounded between \(-1\) and \(1\) (also between \(-1\) and \(\frac12\)). This sequence converges to 0.

  7. This sequence is both bounded and monotone. It's decreasing and bounded between 1 and 2. It converges with limit 1.

Notice in the examples above, being bounded or monotone alone does not guarantee convergence. However, each example that was both bounded and monotone was convergent. This is true in general.

Convergence of Monotone Bounded Sequences

If a sequence \(\{s_n\}\) is bounded and monotone, then it converges.

Boundedness of Convergent Sequences

If a sequence \(\{s_n\}\) converges, then it is bounded.

Besides these two rules Monotone + Bounded \(\rightarrow \) Convergent and Convergent \(\rightarrow \) Bounded), any combination of monotone/not monotone, bounded/not bounded, and convergent/not convergent is possible. For example, we've already seen a sequence in Example7.10 which is bounded but not convergent.

SubsectionSummary

  • A sequence is a list of objects in a specified order. We will typically work with sequences of real numbers. We can think of a sequence as a function from the positive integers to the set of real numbers.

  • A sequence \(\{s_n\}\) of real numbers converges to a number \(L\) if by choosing \(n\) sufficiently large we can make every value of \(s_k\) as close as we want to \(L\text{,}\) for \(k \ge n\text{.}\)

  • A sequence diverges if it does not converge.

  • Sequences are monotone if they are always increasing or always decreasing, and bounded if the terms of the sequence are always between an upper bound and a lower bound. Bounded monotone sequences converge, and convergent sequences are bounded.

SubsectionExercises

Finding limits of convergent sequences can be a challenge. However, there is a useful tool we can adapt from our study of limits of continuous functions at infinity to use to find limits of sequences. We illustrate in this exercise with the example of the sequence

\begin{equation*} \frac{\ln(n)}{n}\text{.} \end{equation*}
  1. Calculate the first 10 terms of this sequence. Based on these calculations, do you think the sequence converges or diverges? Why?

  2. For this sequence, there is a corresponding continuous function \(f\) defined by

    \begin{equation*} f(x) = \frac{\ln(x)}{x}\text{.} \end{equation*}

    Draw the graph of \(f(x)\) on the interval \([0,10]\) and then plot the entries of the sequence on the graph. What conclusion do you think we can draw about the sequence \(\left\{\frac{\ln(n)}{n}\right\}\) if \(\lim_{x \to \infty} f(x) = L\text{?}\) Explain.

  3. Note that \(f(x)\) has the indeterminate form \(\frac{\infty}{\infty}\) as \(x\) goes to infinity. What idea from differential calculus can we use to calculate \(\lim_{x \to \infty} f(x)\text{?}\) Use this method to find \(\lim_{x \to \infty} f(x)\text{.}\) What, then, is \(\lim_{n \to \infty} \frac{\ln(n)}{n}\text{?}\)

We return to the example begun in Preview Activity7.2 to see how to derive the formula for the amount of money in an account at a given time. We do this in a general setting. Suppose you invest \(P\) dollars (called the principal) in an account paying \(r\%\) interest compounded monthly. In the first month you will receive \(\frac{r}{12}\) (here \(r\) is in decimal form; e.g., if we have \(8\%\) interest, we write \(\frac{0.08}{12}\)) of the principal \(P\) in interest, so you earn

\begin{equation*} P\left(\frac{r}{12}\right) \end{equation*}

dollars in interest. Assume that you reinvest all interest. Then at the end of the first month your account will contain the original principal \(P\) plus the interest, or a total of

\begin{equation*} P_1 = P + P\left(\frac{r}{12}\right) = P\left( 1 + \frac{r}{12}\right) \end{equation*}

dollars.

  1. Given that your principal is now \(P_1\) dollars, how much interest will you earn in the second month? If \(P_2\) is the total amount of money in your account at the end of the second month, explain why

    \begin{equation*} P_2 = P_1\left( 1 + \frac{r}{12}\right) = P\left( 1 + \frac{r}{12}\right)^2\text{.} \end{equation*}
  2. Find a formula for \(P_3\text{,}\) the total amount of money in the account at the end of the third month in terms of the original investment \(P\text{.}\)

  3. There is a pattern to these calculations. Let \(P_n\) the total amount of money in the account at the end of the third month in terms of the original investment \(P\text{.}\) Find a formula for \(P_n\text{.}\)

Sequences have many applications in mathematics and the sciences. In a recent paper5Hui H, Farilla L, Merkel P, Perfetti R. The short half-life of glucagon-like peptide-1 in plasma does not reflect its long-lasting beneficial effects, Eur J Endocrinol 2002 Jun;146(6):863-9. the authors write

The incretin hormone glucagon-like peptide-1 (GLP-1) is capable of ameliorating glucose-dependent insulin secretion in subjects with diabetes. However, its very short half-life (1.5-5 min) in plasma represents a major limitation for its use in the clinical setting.

The half-life of GLP-1 is the time it takes for half of the hormone to decay in its medium. For this exercise, assume the half-life of GLP-1 is 5 minutes. So if \(A\) is the amount of GLP-1 in plasma at some time \(t\text{,}\) then only \(\frac{A}{2}\) of the hormone will be present after \(t+5\) minutes. Suppose \(A_0 = 100\) grams of the hormone are initially present in plasma.

  1. Let \(A_1\) be the amount of GLP-1 present after 5 minutes. Find the value of \(A_1\text{.}\)

  2. Let \(A_2\) be the amount of GLP-1 present after 10 minutes. Find the value of \(A_2\text{.}\)

  3. Let \(A_3\) be the amount of GLP-1 present after 15 minutes. Find the value of \(A_3\text{.}\)

  4. Let \(A_4\) be the amount of GLP-1 present after 20 minutes. Find the value of \(A_4\text{.}\)

  5. Let \(A_n\) be the amount of GLP-1 present after \(5n\) minutes. Find a formula for \(A_n\text{.}\)

  6. Does the sequence \(\{A_n\}\) converge or diverge? If the sequence converges, find its limit and explain why this value makes sense in the context of this problem.

  7. Determine the number of minutes it takes until the amount of GLP-1 in plasma is 1 gram.

Continuous data is the basis for analog information, like music stored on old cassette tapes or vinyl records. A digital signal like on a CD or MP3 file is obtained by sampling an analog signal at some regular time interval and storing that information. For example, the sampling rate of a compact disk is 44,100 samples per second. So a digital recording is only an approximation of the actual analog information. Digital information can be manipulated in many useful ways that allow for, among other things, noisy signals to be cleaned up and large collections of information to be compressed and stored in much smaller space. While we won't investigate these techniques in this chapter, this exercise is intended to give an idea of the importance of discrete (digital) techniques.

Let \(f\) be the continuous function defined by \(f(x) = \sin(4x)\) on the interval \([0,10]\text{.}\) A graph of \(f\) is shown in Figure7.11.

Figure7.11The graph of \(f(x) = \sin(4x)\) on the interval \([0,10]\)

We approximate \(f\) by sampling, that is by partitioning the interval \([0,10]\) into uniform subintervals and recording the values of \(f\) at the endpoints.

  1. Ineffective sampling can lead to several problems in reproducing the original signal. As an example, partition the interval \([0,10]\) into 8 equal length subintervals and create a list of points (the sample) using the endpoints of each subinterval. Plot your sample on graph of \(f\) in Figure Figure7.11. What can you say about the period of your sample as compared to the period of the original function?

  2. The sampling rate is the number of samples of a signal taken per second. As the part (a) illustrates, sampling at too small a rate can cause serious problems with reproducing the original signal (this problem of inefficient sampling leading to an inaccurate approximation is called aliasing). There is an elegant theorem called the Nyquist-Shannon Sampling Theorem that says that human perception is limited, which allows that replacement of a continuous signal with a digital one without any perceived loss of information. This theorem also provides the lowest rate at which a signal can be sampled (called the Nyquist rate) without such a loss of information. The theorem states that we should sample at double the maximum desired frequency so that every cycle of the original signal will be sampled at at least two points. Recall that the frequency of a sinusoidal function is the reciprocal of the period. Identify the frequency of the function \(f\) and determine the number of partitions of the interval \([0,10]\) that give us the Nyquist rate.

  3. Humans cannot typically pick up signals above 20 kHz. Explain why, then, that information on a compact disk is sampled at 44,100 Hz.

For each of the sequences below, decide if the sequence is bounded and if it is monotone. (It may help you to list out the first few terms of the sequence.)

  1. \(a_n= 7- (\frac 1 5)^n\)
  2. \(b_n= 4\sin(n)\)
  3. \(c_n= n (-1)^n\)
  4. \(d_n= n^2+n\)

For each of the recursive sequences below, list the first 6 terms, then write an explicit expression for the \(n\)th term. Check your work by looking up the sequence on the Online Encyclopedia of Integer Sequences at http://oeis.org/.

  1. \(a_n= a_{n-1}+3\) for \(n>1\) and \(a_1=4\text{.}\)

  2. \(b_n= nb_{n-1}\) for \(n>1\) and \(b_1=1\text{.}\)

  3. \(c_n= \frac 1 2 c_{n-1}\) for \(n>1\) and \(c_1=5\text{.}\)

  4. \(d_n= d_{n-1}+4n-2\) for \(n>1\) and \(d_1=2\text{.}\)