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3.1: Sequences

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    In the discussion above we used the term “sequence” without giving it a precise mathematical meaning. Let us rectify this now.

    Definition 3.1.1

    A sequence is a list of infinitely 1 many numbers with a specified order. It is denoted

    \[ \big\{a_1,\ a_2,\ a_3,\ \cdots,\ a_n,\ \cdots\big\} \quad\text{or}\quad \big\{a_n\big\} \quad\text{or}\quad \big\{a_n\big\}_{n=1}^\infty \nonumber \]

    We will often specify a sequence by writing it more explicitly, like

    \begin{gather*} \Big\{ a_n = f(n) \Big\}_{n=1}^\infty \end{gather*}

    where \(f(n)\) is some function from the natural numbers to the real numbers.

    Example 3.1.2 Three sequences and another one

    Here are three sequences.

    \begin{align*} &\Big\{1,\ \frac{1}{2},\ \frac{1}{3},\ \cdots,\ \frac{1}{n},\ \cdots\Big\} &&\text{or} &&\Big\{a_n=\frac{1}{n}\Big\}_{n=1}^\infty\\ &\Big\{1,\ 2,\ 3,\ \cdots,\ n,\ \cdots\Big\} &&\text{or} &&\Big\{a_n=n\Big\}_{n=1}^\infty\\ &\Big\{1,\ -1,\ 1,\ -1,\ \cdots,\ (-1)^{n-1},\ \cdots\Big\} &&\text{or} &&\Big\{a_n=(-1)^{n-1}\Big\}_{n=1}^\infty \end{align*}

    It is not necessary that there be a simple explicit formula for the \(n^{\rm th}\) term of a sequence. For example the decimal digits of \(\pi\) is a perfectly good sequence

    \[ \big\{3,\ 1,\ 4,\ 1,\ 5,\ 9,\ 2,\ 6,\ 5,\ 3,\ 5,\ 8,\ 9,\ 7,\ 9,\ 3,\ 2,\ 3,\ 8,\ 4,\ 6,\ 2,\ 6,\ 4,\ \cdots\ \big\} \nonumber \]

    but there is no simple formula 2 for the \(n^{\rm th}\) digit.

    Our primary concern with sequences will be the behaviour of \(a_n\) as \(n\) tends to infinity and, in particular, whether or not \(a_n\) “settles down” to some value as \(n\) tends to infinity.

    Definition 3.1.3

    A sequence \(\big\{a_n\big\}_{n=1}^\infty\) is said to converge to the limit \(A\) if \(a_n\) approaches \(A\) as \(n\) tends to infinity. If so, we write

    \[ \lim_{n\rightarrow\infty} a_n=A\qquad\hbox{or}\qquad a_n\rightarrow A\text{ as }n\rightarrow\infty \nonumber \]

    A sequence is said to converge if it converges to some limit. Otherwise it is said to diverge.

    The reader should immediately recognise the similarity with limits at infinity

    \begin{gather*} \lim_{x \to \infty} f(x) = L \qquad\hbox{if}\qquad f(x) \to L \text{ as } x \to \infty \end{gather*}

    Example 3.1.4 Convergence in Example 3.1.2

    Three of the four sequences in Example 3.1.2 diverge:

    • The sequence \(\big\{a_n=n\big\}_{n=1}^\infty\) diverges because \(a_n\) grows without bound, rather than approaching some finite value, as \(n\) tends to infinity.
    • The sequence \(\big\{a_n=(-1)^{n-1}\big\}_{n=1}^\infty\) diverges because \(a_n\) oscillates between \(+1\) and \(-1\) rather than approaching a single value as \(n\) tends to infinity.
    • The sequence of the decimal digits of \(\pi\) also diverges, though the proof that this is the case is a bit beyond us right now 3.

    The other sequence in Example 3.1.2 has \(a_n=\frac{1}{n}\text{.}\) As \(n\) tends to infinity, \(\frac{1}{n}\) tends to zero. So

    \[ \lim_{n\rightarrow\infty} \frac{1}{n}=0 \nonumber \]

    Example 3.1.5 \(\lim\limits_{n\rightarrow\infty}\frac{n}{2n+1}\)

    Here is a little less trivial example. To study the behaviour of \(\frac{n}{2n+1}\) as \(n\rightarrow\infty\text{,}\) it is a good idea to write it as

    \[ \frac{n}{2n+1}=\frac{1}{2+\frac{1}{n}} \nonumber \]

    As \(n\rightarrow\infty\text{,}\) the \(\frac{1}{n}\) in the denominator tends to zero, so that the denominator \(2+\frac{1}{n}\) tends to \(2\) and \(\frac{1}{2+\frac{1}{n}}\) tends to \(\frac{1}{2}\text{.}\) So

    \begin{gather*} \lim_{n\rightarrow\infty}\frac{n}{2n+1} =\lim_{n\rightarrow\infty}\frac{1}{2+\frac{1}{n}} =\frac{1}{2} \end{gather*}

    Notice that in this last example, we are really using techniques that we used before to study infinite limits like \(\displaystyle \lim_{x\rightarrow\infty}f(x)\text{.}\) This experience can be easily transferred to dealing with \(\lim\limits_{n\rightarrow\infty}a_n\) limits by using the following result.

    Theorem 3.1.6


    \[ \lim_{x\rightarrow\infty} f(x) = L \nonumber \]

    and if \(a_n=f(n)\) for all positive integers \(n\text{,}\) then

    \[ \lim_{n\rightarrow\infty} a_n = L \nonumber \]

    Example 3.1.7 \(\lim\limits_{n\rightarrow\infty}e^{-n}\)

    Set \(f(x)=e^{-x}\text{.}\) Then \(e^{-n}=f(n)\) and

    \begin{align*} \text{since } \lim_{x\rightarrow\infty}e^{-x}&=0 &\text{ we know that }&& \lim\limits_{n\rightarrow\infty}e^{-n}&=0 \end{align*}

    The bulk of the rules for the arithmetic of limits of functions that you already know also apply to the limits of sequences. That is, the rules you learned to work with limits such as \(\displaystyle \lim_{x\rightarrow\infty}f(x)\) also apply to limits like \(\displaystyle \lim_{n\rightarrow\infty}a_n\text{.}\)

    Theorem 3.1.8 Arithmetic of limits

    Let \(A\text{,}\) \(B\) and \(C\) be real numbers and let the two sequences \(\big\{a_n\big\}_{n=1}^\infty\) and \(\big\{b_n\big\}_{n=1}^\infty\) converge to \(A\) and \(B\) respectively. That is, assume that

    \begin{align*} \lim_{n \to \infty} a_n&=A & \lim_{n \to \infty} b_n &=B \end{align*}

    Then the following limits hold.

    1. \(\displaystyle \lim_{n \to \infty} \big[a_n+b_n\big] = A+B\)

      (The limit of the sum is the sum of the limits.)

    2. \(\displaystyle \lim_{n \to \infty} \big[a_n-b_n\big] = A-B\)

      (The limit of the difference is the difference of the limits.)

    3. \(\displaystyle \lim_{n \to \infty} C a_n = C A\text{.}\)
    4. \(\displaystyle \lim_{n \to \infty} a_n\,b_n = A\,B\)

      (The limit of the product is the product of the limits.)

    5. If \(B \neq 0\) then \(\displaystyle \lim_{n \to \infty}\frac{a_n}{b_n} = \frac{A}{B}\)

      (The limit of the quotient is the quotient of the limits provided the limit of the denominator is not zero.)

    We use these rules to evaluate limits of more complicated sequences in terms of the limits of simpler sequences — just as we did for limits of functions.

    Example 3.1.9 Arithmetic of limits

    Combining Examples 3.1.5 and 3.1.7,

    \[\begin{align*} \lim_{n\rightarrow\infty}\Big[\frac{n}{2n+1} + 7 e^{-n}\Big] &= \lim_{n\rightarrow\infty}\frac{n}{2n+1} +\lim_{n\rightarrow\infty} 7 e^{-n} & \text{by Theorem }{\text{3.1.8}}\text{(a)}\\ &= \lim_{n\rightarrow\infty}\frac{n}{2n+1} +7\lim_{n\rightarrow\infty} e^{-n} & \text{by Theorem }{\text{3.1.8}}\text{(c)}\\ \end{align*}\]

    and then using Examples 3.1.5 and 3.1.7

    \begin{align*} &=\frac{1}{2} + 7\cdot 0\\ &=\frac{1}{2} \end{align*}

    There is also a squeeze theorem for sequences.

    Theorem 3.1.10 Squeeze theorem

    If \(a_n\le c_n\le b_n\) for all natural numbers \(n\text{,}\) and if

    \[ \lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=L \nonumber \]


    \[ \lim_{n\rightarrow\infty}c_n=L \nonumber \]

    Example 3.1.11 A simple squeeze

    In this example we use the squeeze theorem to evaluate

    \[ \lim_{n\rightarrow\infty}\Big[1+\frac{\pi_n}{n}\Big] \nonumber \]

    where \(\pi_n\) is the \(n^{\mathrm{th}}\) decimal digit of \(\pi\text{.}\) That is,

    \[ \pi_1=3\quad \pi_2=1 \quad \pi_3=4 \quad \pi_4=1 \quad \pi_5=5 \quad\pi_6=9\quad\cdots \nonumber \]

    We do not have a simple formula for \(\pi_n\text{.}\) But we do know that

    \[ 0\le\pi_n\le 9 \implies 0 \le \frac{\pi_n}{n} \le \frac{9}{n} \implies 1 \le 1+\frac{\pi_n}{n} \le 1+\frac{9}{n} \nonumber \]

    and we also know that

    \begin{gather*} \lim_{n\rightarrow\infty} 1 = 1\qquad \lim_{n\rightarrow\infty} \Big[1+\frac{9}{n}\Big] = 1 \end{gather*}

    So the squeeze theorem with \(a_n=1\text{,}\) \(b_n=1+\frac{\pi_n}{n}\text{,}\) and \(c_n=1+\frac{9}{n}\) gives

    \[ \lim_{n\rightarrow\infty}\Big[1+\frac{\pi_n}{n}\Big] = 1 \nonumber \]

    Finally, recall that we can compute the limit of the composition of two functions using continuity. In the same way, we have the following result:

    Theorem 3.1.12 Continuous functions of limits

    If \(\lim\limits_{n\rightarrow\infty}a_n=L \) and if the function \(g(x)\) is continuous at \(L\text{,}\) then

    \[ \lim_{n\rightarrow\infty}g(a_n)=g(L) \nonumber \]

    Example 3.1.13 \(\lim\limits_{n\rightarrow\infty}\sin\frac{\pi n}{2n+1}\)

    Write \(\sin\frac{\pi n}{2n+1}=g\big(\frac{n}{2n+1}\big)\) with \(g(x)=\sin(\pi x)\text{.}\) We saw, in Example 3.1.5 that

    \[ \lim_{n\rightarrow\infty}\frac{n}{2n+1} = \frac{1}{2} \nonumber \]

    Since \(g(x) = \sin (\pi x)\) is continuous at \(x=\frac{1}{2}\text{,}\) which is the limit of \(\frac{n}{2n+1}\text{,}\) we have

    \[ \lim_{n\rightarrow\infty}\sin\frac{\pi n}{2n+1} =\lim_{n\rightarrow\infty}g\Big(\frac{n}{2n+1}\Big) =g\Big(\frac{1}{2}\Big) =\sin\frac{\pi}{2} =1 \nonumber \]

    With this introduction to sequences and some tools to determine their limits, we can now return to the problem of understanding infinite sums.


    Stage 1

    Assuming the sequences continue as shown, estimate the limit of each sequence from its graph.



    Suppose \(a_n\) and \(b_n\) are sequences, and \(a_n=b_n\) for all \(n \geq 100\text{,}\) but \(a_n \neq b_n\) for \(n \lt 100\text{.}\)

    True or false: \(\displaystyle\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n\text{.}\)


    Let \(\{a_n\}_{n=1}^{\infty}\text{,}\) \(\{b_n\}_{n=1}^{\infty}\text{,}\) and \(\{c_n\}_{n=1}^{\infty}\text{,}\) be sequences with \(\lim\limits_{n \to \infty}a_n=A\text{,}\) \(\lim\limits_{n \to \infty}b_n=B\text{,}\) and \(\lim\limits_{n \to \infty}c_n=C\text{.}\) Assume \(A\text{,}\) \(B\text{,}\) and \(C\) are nonzero real numbers.

    Evaluate the limits of the following sequences.

    1. \(\dfrac{a_n-b_n}{c_n}\)
    2. \(\dfrac{c_n}{n}\)
    3. \(\dfrac{a_{2n+5}}{b_n}\)

    Give an example of a sequence \(\{a_n\}_{n=1}^{\infty}\) with the following properties:

    • \(a_n \gt 1000\) for all \(n \leq 1000\text{,}\)
    • \(a_{n+1} \lt a_n\) for all \(n\text{,}\) and
    • \(\lim\limits_{n \to \infty} a_n = -2\)

    Give an example of a sequence \(\{a_n\}_{n=1}^{\infty}\) with the following properties:

    • \(a_n \gt 0\) for all even \(n\text{,}\)
    • \(a_n \lt 0\) for all odd \(n\text{,}\)
    • \(\lim\limits_{n \to \infty} a_n\) does not exist.

    Give an example of a sequence \(\{a_n\}_{n=1}^{\infty}\) with the following properties:

    • \(a_n \gt 0\) for all even \(n\text{,}\)
    • \(a_n \lt 0\) for all odd \(n\text{,}\)
    • \(\lim\limits_{n \to \infty} a_n\) exists.

    The limits of the sequences below can be evaluated using the squeeze theorem. For each sequence, choose an upper bounding sequence and lower bounding sequence that will work with the squeeze theorem.

    1. \(a_n = \dfrac{\sin n}{n}\)
    2. \(b_n = \dfrac{n^2}{e^n(7+\sin n - 5\cos n)}\)
    3. \(c_n = (-n)^{-n}\)

    Below is a list of sequences, and a list of functions.

    1. Match each sequence \(a_n\) to any and all functions \(f(x)\) such that \(f(n)=a_n\) for all whole numbers \(n\text{.}\)
    2. Match each sequence \(a_n\) to any and all functions \(f(x)\) such that \(\displaystyle\lim_{n \to \infty}a_n = \lim_{x \to \infty}f(x)\text{.}\)

    \begin{align*} a_n &= 1+\dfrac{1}{n} & f(x) &= \cos(\pi x)\\ b_n &= 1+\dfrac{1}{|n|} & g(x) &= \dfrac{\cos (\pi x)}{x}\\ c_n&=e^{-n} & h(x)&=\begin{cases}\frac{x+1}{x}&x\text{ is a whole number}\\ 1&\text{else} \end{cases}\\ d_n&=(-1)^n & i(x)&=\begin{cases}\frac{x+1}{x}&x\text{ is a whole number}\\ 0 & \text{else}\end{cases}\\ e_n&=\dfrac{(-1)^n}{n} & j(x)&=\dfrac{1}{e^x} \end{align*}


    Let \(\{a_n\}_{n=1}^\infty\) be a sequence defined by \(a_n = \cos n\text{.}\)

    1. Give three different whole numbers \(n\) that are within 0.1 of an odd integer multiple of \(\pi\text{,}\) and find the corresponding values of \(a_n\text{.}\)
    2. Give three different whole numbers \(n\) such that \(a_n\) is close to \(0\text{.}\) Justify your answers.

    Remark: this demonstrates intuitively, though not rigorously, why \(\lim\limits_{n \to \infty}\cos n\) is undefined. We consistently find terms in the series that are close to \(-1\text{,}\) and also consistently find terms in the series that are close to 1. Contrast this to a series like \(\big\{\cos(2\pi n)\big\}\text{,}\) whose terms are always 1, and whose limit therefore is 1. It is possible to turn the ideas of this question into a rigorous proof that \(\lim\limits_{n \to \infty}\cos n\) is undefined. See the solution.

    Stage 2

    Determine the limits of the following sequences.

    1. \(a_n = \dfrac{3n^2-2n+5}{4n+3}\)
    2. \(b_n = \dfrac{3n^2-2n+5}{4n^2+3}\)
    3. \(c_n = \dfrac{3n^2-2n+5}{4n^3+3}\)

    Determine the limit of the sequence \(a_n = \dfrac{4n^3-21}{n^e+\frac{1}{n}}\text{.}\)


    Determine the limit of the sequence \(b_n = \dfrac{\sqrt[4]{n}+1}{\sqrt{9n+3}}\text{.}\)


    Determine the limit of the sequence \(c_n = \dfrac{\cos(n+n^2)}{n}\text{.}\)


    Determine the limit of the sequence \(a_n = \dfrac{n^{\sin n}}{n^2}\text{.}\)


    Determine the limit of the sequence \(d_n = e^{-1/n}\text{.}\)


    Determine the limit of the sequence \(a_n = \dfrac{1+3\sin(n^2)-2\sin n}{n}\text{.}\)


    Determine the limit of the sequence \(b_n=\dfrac{e^n}{2^n+n^2}\text{.}\)

    18 (✳)

    Find the limit, if it exists, of the sequence \(\big\{a_k\big\}\text{,}\) where

    \[ a_k=\frac{k!\sin^3 k}{(k+1)!} \nonumber \]

    19 (✳)

    Consider the sequence \(\Big\{(-1)^n\sin\big(\frac{1}{n}\big)\Big\}\text{.}\) State whether this sequence converges or diverges, and if it converges give its limit.

    20 (✳)

    Evaluate \(\displaystyle\lim_{n\rightarrow\infty}\left[\frac{6n^2+5n}{n^2+1} +3\cos(1/n^2) \right] \text{.}\)

    Stage 3
    21 (✳)

    Find the limit of the sequence \(\displaystyle\left\{\log\left(\sin\frac{1}{n}\right)+\log(2n)\right\} \text{.}\)


    Evaluate \(\displaystyle\lim_{n \to \infty}\left[\sqrt{n^2+5n}-\sqrt{n^2-5n}\right]\text{.}\)


    Evaluate \(\displaystyle\lim_{n \to \infty}\left[\sqrt{n^2+5n}-\sqrt{2n^2-5}\right]\text{.}\)


    Evaluate the limit of the sequence \(\left\{n\left[\left(2+\frac1n\right)^{100}-2^{100}\right]\right\}_{n=1}^{\infty}\text{.}\)


    Write a sequence \(\{a_n\}_{n=1}^\infty\) whose limit is \(f'(a)\) for a function \(f(x)\) that is differentiable at the point \(a\text{.}\)

    Your answer will depend on \(f\) and \(a\text{.}\)


    Let \(\{A_n\}_{n=3}^\infty\) be the area of a regular polygon with \(n\) sides, with the distance from the centroid of the polygon to each corner equal to 1.


    1. By dividing the polygon into \(n\) triangles, give a formula for \(A_n\text{.}\)
    2. What is \(\lim\limits_{n \to \infty} A_n\text{?}\)

    Suppose we define a sequence \(\{f_n\}\text{,}\) which depends on some constant \(x\text{,}\) as the following:

    \[ f_n(x) = \begin{cases} 1 & n \leq x \lt n+1\\ 0 & \text{else} \end{cases} \nonumber \]

    For a fixed constant \(x \ge 1\text{,}\) \(\{f_n\}\) is the sequence \(\{0,0,0,\ldots,0,1,0,\ldots,0,0,0,\ldots\}\text{.}\) The sole nonzero element comes in position \(k\text{,}\) where \(k\) is what we get when we round \(x\) down to a whole number. If \(x \lt 1\text{,}\) then the sequence consists of all zeroes.

    Since we can plug in different values of \(x\text{,}\) we can think of \(f_n(x)\) as a function of sequences: a different \(x\) gives you a different sequence. On the other hand, if we imagine fixing \(n\text{,}\) then \(f_n(x)\) is just a function, where \(f_n(x)\) gives the \(n\)th term in the sequence corresponding to \(x\text{.}\)

    1. Sketch the curve \(y=f_2(x)\text{.}\)
    2. Sketch the curve \(y=f_3(x)\text{.}\)
    3. Define \(A_n = \int_0^\infty f_n(x)\,\, d{x}\text{.}\) Give a simple description of the sequence \(\{A_n\}_{n=1}^\infty\text{.}\)
    4. Evaluate \(\displaystyle\lim_{n \to \infty}A_n\text{.}\)
    5. Evaluate \(\displaystyle\lim_{n \to \infty} f_n(x)\) for a constant \(x\text{,}\) and call the result \(g(x)\text{.}\)
    6. Evaluate \(\displaystyle \int_0^\infty g(x)\,\, d{x}\text{.}\)

    Determine the limit of the sequence \(\displaystyle b_n=\left(1+\frac{3}{n}+\frac{5}{n^2}\right)^n\text{.}\)


    A sequence \(\big\{a_n\big\}_{n=1}^\infty\) of real numbers satisfies the recursion relation \(a_{n+1} = \dfrac{a_n+8}{3}\) for \(n\ge 2\text{.}\)

    1. Suppose \(a_1=4\text{.}\) What is \(\lim\limits_{n \to \infty}a_n\text{?}\)
    2. Find \(x\) if \(x=\dfrac{x+8}{3}\text{.}\)
    3. Suppose \(a_1=1\text{.}\) Show that \(\lim\limits_{n\rightarrow\infty} a_n = L\text{,}\) where \(L\) is the solution to equation above.

    Zipf's Law applied to word frequency can be phrased as follows:

    The most-used word in a language is used \(n\) times as frequently as the \(n\)-th most word used in a language.

    1. Suppose the sequence \(\{w_1,w_2,w_3,\ldots\}\) is a list of all words in a language, where \(w_n\) is the word that is the \(n\)th most frequently used. Let \(f_n\) be the frequency of word \(w_n\text{.}\) Is \(\{f_1,f_2,f_3,\ldots\}\) an increasing sequence or a decreasing sequence?
    2. Give a general formula for \(f_n\text{,}\) treating \(f_1\) as a constant.
    3. Suppose in a language, \(w_1\) (the most frequently used word) has frequency \(6\%\text{.}\) If the language follows Zipf's Law, then what frequency does \(w_3\) have?
    4. Suppose \(f_6=0.3\%\) for a language following Zipf's law. What is \(f_{10}\text{?}\)
    5. The word “the” is the most-used word in contemporary American English. In a collection of about 450 million words, “the” appeared 22,038,615 times. The second-most used word is “be,” followed by “and.” About how many usages of these words do you expect in the same collection of 450 million words?


    • Zipf’s word frequency law in natural language: A critical review and future directions, Steven T. Piantadosi. Psychon Bull Rev. 2014 Oct; 21(5): 1112–1130. Accessed online 11 October 2017.
    • Word Frequency Data. Accessed online 11 October 2017.
    1. For the more pedantic reader, here we mean a countably infinite list of numbers. The interested (pedantic or otherwise) reader should look up countable and uncountable sets.
    2. There is, however, a remarkable result due to Bailey, Borwein and Plouffe that can be used to compute the \(n^{\rm th}\) binary digit of \(\pi\) (i.e. writing \(\pi\) in base 2 rather than base 10) without having to work out the preceding digits.
    3. If the digits of \(\pi\) were to converge, then \(\pi\) would have to be a rational number. The irrationality of \(\pi\) (that it cannot be written as a fraction) was first proved by Lambert in 1761. Niven's 1947 proof is more accessible and we invite the interested reader to use their favourite search engine to find step-by-step guides to that proof.

    This page titled 3.1: Sequences is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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