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Mathematics LibreTexts

2.1: Sequences

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    22643
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    Definition

    Let \(\left\{a_{i}\right\}_{i \in I}\) be a sequence of real numbers. We say \(\left\{a_{i}\right\}_{i \in I}\) converges, and has limit \(L,\) if for every real number \(\epsilon>0\) there exists an integer \(N\) such that

    \[\left|a_{i}-L\right|<\epsilon \text { whenever } i>N.\]

    We say a sequence \(\left\{a_{i}\right\}_{i \in I}\) which does not converge diverges.

    Definitions: nondecreasing and nonincreasing sequences
    • We say a sequence \(\left\{a_{i}\right\}_{i \in I}\) is nondecreasing if \(a_{i+1} \geq a_{i}\) for every \(i \in I\) and increasing if \(a_{i+1}>a_{i}\) for every \(i \in I \).
    • We say a sequence \(\left\{a_{i}\right\}_{i \in I}\) is nonincreasing if \(a_{i+1} \leq a_{i}\) for every \(i \in I\) and decreasing if \(a_{i+1}<a_{i}\) for every \(i \in I \).
    Definition: bounded sequences

    We say a set \(A \subset \mathbb{R}\) is bounded if there exists a real number \(M\) such that \(|a| \leq M\) for every \(a \in A .\) We say a sequence \(\left\{a_{i}\right\}_{i \in I}\) of real numbers is bounded if there exists a real number \(M\) such that \(\left|a_{i}\right| \leq M\) for all \(i \in I .\)

    Theorem \(\PageIndex{1}\)

    If \(\left\{a_{i}\right\}_{i \in I}\) is a nondecreasing, bounded sequence of real numbers, then \(\left\{a_{i}\right\}_{i \in I}\) converges.

    Proof

    Since \(\left\{a_{i}\right\}_{i \in I}\) is bounded, the set of \(A=\left\{a_{i}: i \in I\right\}\) has a supremum. Let \(L=\sup A .\) For any \(\epsilon>0,\) there must exist \(N \in I\) such that \(a_{N}>L-\epsilon\) (or else \(L-\epsilon\) would be an upper bound for \(A\) which is smaller than \(L\) ). But then

    \[L-\epsilon<a_{N} \leq a_{i} \leq L<L+\epsilon\]

    for all \(i \geq N,\) that is,

    \[\left|a_{i}-L\right|<\epsilon\]

    for all \(i \geq N .\) Thus \(\left\{a_{i}\right\}_{i \in I}\) converges and

    \[L=\lim _{i \rightarrow \infty} a_{i}.\]

    Q.E.D.

    We conclude from the previous theorem that every nondecreasing sequence of real numbers either has a limit or is not bounded, that is, is unbounded.

    Exercise \(\PageIndex{1}\)

    Show that a nonincreasing, bounded sequence of real numbers must converge.

    Definition

    Let \(\left\{a_{i}\right\}_{i \in I}\) be a sequence of real numbers. If for every real number \(M\) there exists an integer \(N\) such that \(a_{i}>M\) whenever \(i>N,\) then we say the sequence \(\left\{a_{i}\right\}_{i \in I}\) diverges to positive infinity, denoted by

    \[\lim _{i \rightarrow \infty} a_{i}=+\infty.\]

    Similarly, if for every real number \(M\) there exists an integer \(N\) such that \(a_{i}<M\) whenever \(i>N,\) then we say the sequence \(\left\{a_{i}\right\}_{i \in I}\) diverges to negative infinity, denoted by

    \[\lim _{i \rightarrow \infty} a_{i}=-\infty.\]

    Exercise \(\PageIndex{2}\)

    Show that a nondecreasing sequence of real numbers either converges or diverges to positive infinity.

    Exercise \(\PageIndex{3}\)

    Show that a nonincreasing sequence of real numbers either converges or diverges to negative infinity.

    2.1.1 Extended Real Numbers

    It is convenient to add the symbols \(+\infty\) and \(-\infty\) to the real numbers \(\mathbb{R} .\) Although neither \(+\infty\) nor \(-\infty\) is a real number, we agree to the following operational conventions:

    1. Given any real number \(r,-\infty<r<\infty\).

    2. For any real number \(r\),

    \[r+(+\infty)=r-(-\infty)=r+\infty=+\infty,\]

    \[r+(-\infty)=r-(+\infty)=r-\infty=-\infty,\]

    and

    \[\frac{r}{+\infty}=\frac{r}{-\infty}=0.\]

    3. For any real number \(r>0, r \cdot(+\infty)=+\infty\) and \(r \cdot(-\infty)=-\infty\).

    4. For any real number \(r<0, r \cdot(+\infty)=-\infty\) and \(r \cdot(-\infty)=+\infty\).

    5. If \(a_{i}=-\infty, i=1,2,3, \ldots,\) then \(\lim _{i \rightarrow \infty} a_{i}=-\infty\).

    6. If \(a_{i}=+\infty, i=1,2,3, \ldots,\) then \(\lim _{i \rightarrow \infty} a_{i}=+\infty\).

    Note that with the order relation defined in this manner, \(+\infty\) is an upper

    bound and \(-\infty\) is a lower bound for any set \(A \subset \mathbb{R}\). Thus if \(A \subset \mathbb{R}\) does not have a finite upper bound, then \(\sup A=+\infty ;\) similarly, if \(A \subset \mathbb{R}\) does not have a finite lower bound, then inf \(A=-\infty\).

    When working with extended real numbers, we refer to the elements of \(\mathbb{R}\) as finite real numbers and the elements \(+\infty\) and \(-\infty\) as infinite real numbers.

    Exercise \(\PageIndex{4}\)

    Do the extended real numbers form a field?

    2.1.2 Limit Superior and Inferior

    Definition

    Let \(\left\{a_{i}\right\}_{i \in I}\) be a sequence of real numbers and, for each \(i \in I,\) let \(u_{i}=\sup \left\{a_{j}: j \geq i\right\} .\) If \(u_{i}=+\infty\) for every \(i \in I,\) we let

    \[\limsup _{i \rightarrow \infty} a_{i}=+\infty;\]

    otherwise, we let

    \[\limsup _{i \rightarrow \infty} a_{i}=\inf \left\{u_{i}: i \in I\right\}.\]

    In either case, we call \(\liminf _{n \rightarrow \infty} a_{n}\) the limit inferior of the sequence \(\left\{a_{i}\right\}_{i \in I}\).

    Exercise \(\PageIndex{5}\)

    Given a sequence \(\left\{a_{i}\right\}_{i \in I},\) define \(\left\{u_{i}\right\}_{i \in I}\) and \(\left\{l_{i}\right\}_{i \in I}\) as in the previous two definitions. Show that

    \[\limsup _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} u_{i}\]

    and

    \[\liminf _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} l_{i}.\]

    Exercise \(\PageIndex{6}\)

    Find \(\limsup _{i \rightarrow \infty} a_{i}\) and \(\liminf _{i \rightarrow \infty} a_{i}\) for the sequences \(\left\{a_{i}\right\}_{i=1}^{\infty}\) as de fined below.

    a. \(a_{i}=(-1)^{i}\)

    b. \(a_{i}=i\)

    c. \(a_{i}=2^{-i}\)

    d. \(a_{i}=\frac{1}{i}\)

    The following proposition is often called the squeeze theorem.

    Proposition \(\PageIndex{2}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I},\left\{b_{j}\right\}_{j \in J},\) and \(\left\{c_{k}\right\}_{k \in K}\) are sequences of real numbers for which there exists an integer \(N\) such that \(a_{i} \leq c_{i} \leq b_{i}\) whenever \(i>N .\) If

    \[\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i},\]

    then

    \[\lim _{i \rightarrow \infty} c_{i}=\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i}.\]

    Proof

    Let \(L=\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i} .\) Suppose \(L\) is finite. Given \(\epsilon>0,\) there exists an integer \(M\) such that

    \[\left|a_{i}-L\right|<\frac{\epsilon}{3}\]

    and

    \[\left|b_{i}-L\right|<\frac{\epsilon}{3}\]

    whenever \(i>M .\) Then

    \[\left|a_{i}-b_{i}\right| \leq\left|a_{i}-L\right|+\left|L-b_{i}\right|<\frac{\epsilon}{3}+\frac{\epsilon}{3}=\frac{2 \epsilon}{3}\]

    whenever \(i>M .\) Let \(K\) be the larger of \(N\) and \(M .\) Then

    \[\left|c_{i}-L\right| \leq\left|c_{i}-b_{i}\right|+\left|b_{i}-L\right| \leq\left|a_{i}-b_{i}\right|+\left|b_{i}-L\right|<\frac{2 \epsilon}{3}+\frac{\epsilon}{3}=\epsilon\]

    whenever \(i>K .\) Thus lim \(c_{i}=L .\) The result when \(L\) is infinite is a consequence of the next two exercises. \(\quad\) Q.E.D.

    Exercise \(\PageIndex{7}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) and \(\left\{c_{k}\right\}_{k \in K}\) are sequences for which there exists an integer \(N\) such that \(a_{i} \leq c_{i}\) whenever \(i>N .\) Show that if \(\lim _{i \rightarrow \infty} a_{i}=+\infty,\) then \(\lim _{i \rightarrow \infty} c_{i}=+\infty .\)

    Exercise \(\PageIndex{8}\)

    Suppose \(\left\{b_{j}\right\}_{j \in J}\) and \(\left\{c_{k}\right\}_{k \in K}\) are sequences for which there exists an integer \(N\) such that \(c_{i} \leq b_{i}\) whenever \(i>N .\) Show that if \(\lim _{i \rightarrow \infty} b_{i}=-\infty,\) then \(\lim _{i \rightarrow \infty} c_{i}=-\infty .\)

    Exercise \(\PageIndex{9}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) and \(\left\{b_{j}\right\}_{j \in J}\) are sequences of real numbers with \(a_{i} \leq b_{i}\) for all \(i\) larger than some integer \(N .\) Assuming both sequences converge, show that

    \[\lim _{\mathfrak{i} \rightarrow \infty} a_{i} \leq \lim _{i \rightarrow \infty} b_{i}.\]

    Exercise \(\PageIndex{10}\)

    Show that for any sequence \(\left\{a_{i}\right\}_{i \in I}\),

    \[\liminf _{i \rightarrow \infty} a_{i} \leq \limsup _{i \rightarrow \infty} n_{i}.\]

    Proposition \(\PageIndex{3}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) is a sequence for which

    \[\limsup _{i \rightarrow \infty} a_{i}=\liminf _{i \rightarrow \infty} a_{i}.\]

    Then

    \[\lim _{i \rightarrow \infty} a_{i}=\limsup _{i \rightarrow \infty} a_{i}=\liminf _{i \rightarrow \infty} a_{i}.\]

    Proof

    Let \(u_{i}=\sup \left\{a_{k}: k \geq i\right\}\) and \(l_{i}=\inf \left\{a_{k}: k \geq i\right\} .\) Then \(l_{i} \leq a_{i} \leq u_{i}\) for all \(i \in I .\) Now

    \[\lim _{i \rightarrow \infty} l_{i}=\liminf _{i \rightarrow \infty} a_{i}=\limsup _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} u_{i},\]

    so the result follows from the squeeze theorem. \(\quad\) Q.E.D.

    Exercise \(\PageIndex{11}\)

    Suppose \(u\) is a real number such that \(u \geq 0\) and \(u<\epsilon\) for any real number \(\epsilon>0 .\) Show that \(u=0\).

    2.1.3 Completeness

    Definition

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) is a sequence in \(\mathbb{R} .\) We call \(\left\{a_{i}\right\}_{i \in I}\) a Cauchy sequence if for every \(\epsilon>0\) there exists an integer \(N\) such that

    \[\left|a_{i}-a_{j}\right|<\epsilon\]

    whenever both \(i>N\) and \(j>N\).

    Theorem \(\PageIndex{4}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) is a Cauchy sequence in \(\mathbb{R} .\) Then \(\left\{a_{i}\right\}_{i \in I}\) converges to a limit \(L \in \mathbb{R}\).

    Proof

    Let \(u_{i}=\sup \left\{a_{k}: k \geq i\right\}\) and \(l_{i}=\inf \left\{a_{k}: k \geq i\right\} .\) Given any \(\epsilon>0\), there exists \(N \in \mathbb{Z}\) such that \(\left|a_{i}-a_{j}\right|<\epsilon\) for all \(i, j>N .\) Thus, for all \(i, j>N\), \(a_{i}<a_{j}+\epsilon_{1}\) and so

    \[a_{i} \leq \inf \left\{a_{j}+\epsilon: j \geq i\right\}=l_{i}+\epsilon\]

    for all \(i>N .\) Since \(\left\{l_{i}\right\}_{i \in I}\) is a nondecreasing sequence,

    \[a_{i} \leq \sup \left\{l_{i}+\epsilon: i \in I\right\}=\liminf _{i \rightarrow \infty} a_{i}+\epsilon\]

    for all \(i>N .\) Hence

    \[u_{i}=\sup \left\{a_{k}: k \geq i\right\} \leq \liminf _{i \rightarrow \infty} a_{i}+\epsilon\]

    for all \(i>N .\) Thus

    \[\limsup _{i \rightarrow \infty} a_{i}=\inf \left\{u_{i}: i \in I\right\} \leq \liminf _{i \rightarrow \infty} a_{i}+\epsilon.\]

    Since \(\liminf _{i \rightarrow \infty} a_{i} \leq \limsup _{i \rightarrow \infty} a_{i},\) it follows that

    \[\left|\limsup _{i \rightarrow \infty} a_{i}-\liminf _{i \rightarrow \infty} a_{i}\right| \leq \epsilon.\]

    since this is true for every \(\epsilon>0,\) we have \(\lim \sup _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} \inf a_{i},\) and so \(\left\{a_{i}\right\}_{i \in I}\) converges by Proposition \(2.1 .3 .\) \(\quad\) Q.E.D.

    As a consequence of the previous theorem, we say that \(\mathbb{R}\) is a complete metric space.

    Exercise \(\PageIndex{12}\)

    Suppose \(A \subset \mathbb{R}, A \neq \emptyset,\) and \(s=\sup A .\) Show that there exists a sequence \(\left\{a_{i}\right\}_{i=1}^{\infty}\) with \(a_{i} \in A\) such that \(\lim _{i \rightarrow \infty} a_{i}=s\).

    Exercise \(\PageIndex{13}\)

    Given a real number \(x \geq 0,\) show that there exists a real number \(s \geq 0\) such that \(s^{2}=x\).

    We let \(\sqrt{x}\) denote the number \(s\) in the previous exercise, the square root of \(x\).

    2.1.4 Some Basic Results About Sequences
    Proposition \(\PageIndex{5}\)

    Suppose \(\left\{x_{i}\right\}_{i \in I}\) is a convergent sequence in \(\mathbb{R}, \alpha\) is a real number, and \(L=\lim _{i \rightarrow \infty} x_{i} .\) Then the sequence \(\left\{\alpha x_{i}\right\}_{i \in I}\) oonverges and

    \[\lim _{i \rightarrow \infty} \alpha x_{i}=\alpha L.\]

    Proof

    If \(\alpha=0,\) then \(\left\{\alpha x_{i}\right\}_{i \in I}\) clearly converges to \(0 .\) So assume \(\alpha \neq 0 .\) Given \(\epsilon>0,\) choose an integer \(N\) such that

    \[\left|x_{i}-L\right|<\frac{\epsilon}{|\alpha|}\]

    whenever \(i>N .\) Then for any \(i>N\) we have

    \[\left|\alpha x_{i}-\alpha L\right|=|\alpha|\left|x_{i}-L\right|<|\alpha| \frac{\epsilon}{|\alpha|}=\epsilon.\]

    Thus \(\lim _{i \rightarrow \infty} \alpha x_{i}=\alpha L\). \(\quad\) Q.E.D.

    Proposition \(\PageIndex{6}\)

    Suppose \(\left\{x_{i}\right\}_{i \in I}\) and \(\left\{y_{i}\right\}_{i \in I}\) are convergent sequences in \(\mathbb{R}\) with \(L=\lim _{i \rightarrow \infty} x_{i}\) and \(M=\lim _{i \rightarrow \infty} y_{i} .\) Then the sequence \(\left\{x_{i}+y_{i}\right\}_{i \in I}\) converges and

    \[\lim _{i \rightarrow \infty}\left(x_{i}+y_{i}\right)=L+M.\]

    Exercise \(\PageIndex{14}\)

    Prove the previous proposition.

    Proposition \(\PageIndex{7}\)

    Suppose \(\left\{x_{i}\right\}_{i \in I}\) and \(\left\{y_{i}\right\}_{i \in I}\) are convergent sequences in \(\mathbb{R}\) with \(L=\lim _{i \rightarrow \infty} x_{i}\) and \(M=\lim _{i \rightarrow \infty} y_{i} .\) Then the sequence \(\left\{x_{i} y_{i}\right\}_{i \in I}\) converges and

    \[\lim _{i \rightarrow \infty} x_{i} y_{i}=L M.\]

    Exercise \(\PageIndex{15}\)

    Prove the previous proposition.

    Proposition \(\PageIndex{8}\)

    Suppose \(\left\{x_{i}\right\}_{i \in I}\) and \(\left\{y_{i}\right\}_{i \in I}\) are convergent sequences in \(\mathbb{R}\) with \(L=\lim _{i \rightarrow \infty} x_{i}, M=\lim _{i \rightarrow \infty} y_{i},\) and \(y_{i} \neq 0\) for all \(i \in I .\) If \(M \neq 0,\) then the sequence \(\left\{\frac{x_{i}}{y_{i}}\right\}_{i \in I}\) converges and

    \[\lim _{i \rightarrow \infty} \frac{x_{i}}{y_{i}}=\frac{L}{M}.\]

    Proof

    Since \(M \neq 0\) and \(M=\lim _{i \rightarrow \infty} y_{i},\) we may choose an integer \(N\) such that

    \[\left|y_{i}\right|>\frac{|M|}{2}\]

    whenever \(i>N .\) Let \(B\) be an upper bound for \(\left\{\left|x_{i}\right|: i \in I\right\} \cup\left\{\left|y_{i}\right|: i \in I\right\} .\) Given any \(\epsilon>0,\) we may choose an integer \(P\) such that

    \[\left|x_{i}-L\right|<\frac{M^{2} \epsilon}{4 B}\]

    and

    \[\left|y_{i}-M\right|<\frac{M^{2} \epsilon}{4 B}\]

    whenever \(i>P .\) Let \(K\) be the larger of \(N\) and \(P .\) Then, for any \(i>K,\) we have

    \[\begin{aligned}\left|\frac{x_{i}}{y_{i}}-\frac{L}{M}\right| &=\frac{\left|x_{i} M-y_{i} L\right|}{\left|y_{i} M\right|} \\ &=\frac{\left|x_{i} M-x_{i} y_{i}+x_{i} y_{i}-y_{i} L\right|}{\left|y_{i} M\right|} \\ & \leq \frac{\left|x_{i}\right|\left|M-y_{i}\right|+\left|y_{i}\right|\left|x_{i}-L\right|}{\left|y_{i} M\right|} \\ &<\frac{B \frac{M^{2} \epsilon}{4 B}+B \frac{M^{2} \epsilon}{4 B}}{\frac{M^{2}}{2}} \\ &=\epsilon . \end{aligned}\]

    Thus

    \[\lim _{i \rightarrow \infty} \frac{x_{i}}{y_{i}}=\frac{L}{M}.\]

    Q.E.D.

    Exercise \(\PageIndex{16}\)

    a. Show that \(\lim _{n \rightarrow \infty} \frac{1}{n}=0\).

    b. Show that \(\lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0\) by (i) using the definition of limit directly and then (ii) using previous results.

    Exercise \(\PageIndex{17}\)

    Show that for any positive integer \(k\),

    \[\lim _{n \rightarrow \infty} \frac{1}{n^{k}}=0.\]

    Example \(\PageIndex{1}\)

    We may combine the properties of this section to compute

    \[\begin{aligned} \lim _{n \rightarrow \infty} \frac{5 n^{3}+3 n-6}{2 n^{3}+2 n^{2}-7} &=\lim _{n \rightarrow \infty} \frac{5+\frac{3}{n^{2}}-\frac{6}{n^{3}}}{2+\frac{2}{n}-\frac{7}{n^{3}}} \\ &=\frac{\lim _{n \rightarrow \infty} 5+3 \lim _{n \rightarrow \infty} \frac{1}{n^{2}}-6 \lim _{n \rightarrow \infty} \frac{1}{n^{3}}}{\lim _{n \rightarrow \infty} 2+2 \lim _{n \rightarrow \infty} \frac{1}{n}-7 \lim _{n \rightarrow \infty} \frac{1}{n^{3}}} \\ &=\frac{5+0+0}{2+0+0} \\ &=\frac{5}{2}. \end{aligned}\]

    Exercise \(\PageIndex{18}\)

    Evaluate

    \[\lim _{n \rightarrow \infty} \frac{3 n^{5}+8 n^{3}-6 n}{8 n^{5}+2 n^{4}-31},\]

    carefully showing each step.

    Proposition \(\PageIndex{9}\)

    Suppose \(\left\{x_{i}\right\}_{i \in I}\) is a convergent sequence of nonnegative real numbers with \(L=\lim _{i \rightarrow \infty} x_{i} .\) Then the sequence \(\{\sqrt{x_{i}}\}_{i \in I}\) converges and

    \[\lim _{i \rightarrow \infty} \sqrt{x_{i}}=\sqrt{L}.\]

    Proof

    Let \(\epsilon>0\) be given. Suppose \(L>0\) and note that

    \[\left|x_{i}-L\right|=|\sqrt{x_{i}}-\sqrt{L}||\sqrt{x_{i}}+\sqrt{L}|\]

    implies that

    \[|\sqrt{x_{i}}-\sqrt{L}|=\frac{\left|x_{i}-L\right|}{|\sqrt{x_{i}}+\sqrt{L}|}\]

    for any \(i \in I .\) Choose an integer \(N\) such that

    \[\left|x_{i}-L\right|<\sqrt{L} \epsilon\]

    whenever \(i>N .\) Then, for any \(i>N\),

    \[|\sqrt{x_{i}}-\sqrt{L}|=\frac{\left|x_{i}-L\right|}{|\sqrt{x_{i}}+\sqrt{L}|}<\frac{\sqrt{L} \epsilon}{\sqrt{L}}=\epsilon.\]

    Hence \(\lim _{i \rightarrow \infty} \sqrt{x_{i}}=\sqrt{L}\).

    If \(L=0, \lim _{i \rightarrow \infty} x_{i}=0,\) so we may choose an integer \(N\) such that \(\left|x_{i}\right|<\epsilon^{2}\) for all \(i>N .\) Then

    \[|\sqrt{x_{i}}|<\epsilon\]

    whenever \(i>N,\) so \(\lim _{i \rightarrow \infty} \sqrt{x_{i}}=0\). \(\quad\) Q.E.D.

    Exercise \(\PageIndex{19}\)

    Evaluate

    \[\lim _{n \rightarrow \infty} \frac{\sqrt{3 n^{2}+1}}{5 n+6},\]

    carefully showing each step.

    Exercise \(\PageIndex{20}\)

    Given real numbers \(r>0\) and \(\alpha,\) show that \((\mathrm{a}) \alpha r<r\) if \(0<\alpha<1\) and \((\mathrm{b}) r<\alpha r\) if \(\alpha>1\).

    Proposition \(\PageIndex{10}\)

    If \(x \in \mathbb{R}\) and \(|x|<1,\) then

    \[\lim _{n \rightarrow \infty} x^{n}=0.\]

    Proof

    We first assume \(x \geq 0\). Then the sequence \(\left\{x^{n}\right\}_{n=1}^{\infty}\) is nonincreasing and bounded below by 0. Hence the sequence converges. Let \(L=\lim _{n \rightarrow \infty} x^{n} .\) Then

    \[L=\lim _{n \rightarrow \infty} x^{n}=x \lim _{n \rightarrow \infty} x^{n-1}=x L,\]

    from which it follows that \(L(1-x)=0 .\) since \(1-x>0,\) we must have \(L=0\). The result for \(x<0\) follows from the next exercise. \(\quad\) Q.E.D.

    Exercise \(\PageIndex{21}\)

    Show that \(\lim _{n \rightarrow \infty}\left|a_{n}\right|=0\) if and only if \(\lim _{n \rightarrow \infty} a_{n}=0\).

    2.1.5 Subsequences
    Definition

    Given a sequence \(\left\{x_{i}\right\}_{i=m}^{\infty},\) suppose \(\left\{n_{k}\right\}_{k=1}^{\infty}\) is an increasing sequence of integers with

    \[m \leq n_{1}<n_{2}<n_{3}<\cdots .\]

    Then we call the sequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) a subsequence of \(\left\{x_{i}\right\}_{i=m}^{\infty} .\)

    Example \(\PageIndex{2}\)

    The sequence \(\left\{x_{2 k}\right\}_{k=1}^{\infty}\) is a subsequence of the sequence \(\left\{x_{i}\right\}_{i=1}^{\infty} .\) For example, \(\left\{\frac{1}{2 i}\right\}_{i=1}^{\infty}\) is a subsequence of \(\left\{\frac{1}{i}\right\}_{i=1}^{\infty}\).

    Exercise \(\PageIndex{22}\)

    Suppose \(\left\{x_{i}\right\}_{i=m}^{\infty}\) converges with \(\lim _{i \rightarrow \infty} x_{i}=L .\) Show that every subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) of \(\left\{x_{i}\right\}_{i=m}^{\infty}\) also converges and \(\lim _{k \rightarrow \infty} x_{n_{k}}=L\).

    Exercise \(\PageIndex{23}\)

    Suppose \(\left\{x_{i}\right\}_{i=m}^{\infty}\) diverges to \(+\infty .\) Show that every subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) of \(\left\{x_{i}\right\}_{i=m}^{\infty}\) also diverges to \(+\infty\).

    Exercise \(\PageIndex{24}\)

    Suppose \(\left\{x_{i}\right\}_{i=m}^{\infty}\) diverges to \(-\infty .\) Show that every subsequence \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) of \(\left\{x_{i}\right\}_{i=m}^{\infty}\) also diverges to \(-\infty\).

    Definition

    Given a sequence \(\left\{x_{i}\right\}_{i=m}^{\infty},\) we call any extended real number \(\lambda\) which is the limit of a subsequence of \(\left\{x_{i}\right\}_{i=m}^{\infty}\) a subsequential limit of \(\left\{x_{i}\right\}_{i=m}^{\infty}\).

    Example \(\PageIndex{3}\)

    \(-1\) and 1 are both subsequential limits of \(\left\{(-1)^{i}\right\}_{i=0}^{\infty}\).

    Exercise \(\PageIndex{25}\)

    Suppose the sequence \(\left\{x_{i}\right\}_{i=m}^{\infty}\) is not bounded. Show that either \(-\infty\) or \(+\infty\) is a subsequential limit of \(\left\{x_{i}\right\}_{i=m}^{\infty} .\).

    Proposition \(\PageIndex{11}\)

    Suppose \(\Lambda\) is the set of all subsequential limits of the sequence \(\left\{x_{i}\right\}_{i=m}^{\infty} .\) Then \(\Lambda \neq \emptyset\).

    Proof

    By the previous exercise, the proposition is true if \(\left\{x_{i}\right\}_{i=m}^{\infty}\) is not bounded. So suppose \(\left\{x_{i}\right\}_{i=m}^{\infty}\) is bounded and choose real numbers \(a\) and \(b\) such that \(a \leq x_{i} \leq b\) for all \(i \geq m\). Construct sequences \(\left\{a_{i}\right\}_{i=1}^{\infty}\) and \(\left\{b_{i}\right\}_{i=1}^{\infty}\) as follows: Let \(a_{1}=a\) and \(b_{1}=b\). For \(i \geq 1,\) let

    \[c=\frac{a_{i-1}+b_{i-1}}{2}.\]

    If there exists an integer \(N\) such that \(a_{i-1} \leq x_{j} \leq c\) for all \(j>N,\) let \(a_{i}=a_{i-1}\) and \(b_{i}=c ;\) otherwise, let \(a_{i}=c\) and \(b_{i}=b_{i-1} .\) Let \(n_{1}=m\) and, for \(k=2,3,4, \ldots,\) let \(n_{k}\) be the smallest integer for which \(n_{k}>n_{k-1}\) and \(a_{k} \leq x_{n_{k}} \leq b_{k}\) Then \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) is a Cauchy sequence which is a subsequence of \(\left\{x_{i}\right\}_{i=m}^{\infty} .\) Thus \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) converges and \(\Lambda \neq 0 .\) \(\quad\) Q.E.D.

    Exercise \(\PageIndex{26}\)

    Suppose \(A \subset \mathbb{R}\) and \(B \subset \mathbb{R}\) with \(a \leq b\) for every \(a \in A\) and \(b \in B .\) Show that \(\sup A \leq \inf B\).

    Proposition \(\PageIndex{12}\)

    Let \(\Lambda\) be the set of subsequential limits of a sequence \(\left\{x_{i}\right\}_{i=m}^{\infty} .\) Then

    \[\limsup _{i \rightarrow \infty} x_{i}=\sup \Lambda .\]

    Proof

    Let \(s=\sup \Lambda\) and, for \(i \geq m, u_{i}=\sup \left\{x_{j}: j \geq i\right\} .\) Now since \(x_{j} \leq u_{i}\) for all \(j \geq i,\) it follows that \(\lambda \leq u_{i}\) for every \(\lambda \in \Lambda\) and \(i \geq m .\) Hence, from the previous exercise, \(s \leq \inf \left\{u_{i}: i \geq m\right\}=\limsup _{i \rightarrow \infty} x_{i}\).

    Now suppose \(s<\limsup _{i \rightarrow \infty} x_{i}\). Then there exists a real number \(t\) such that \(s<t<\lim \sup _{i \rightarrow \infty} x_{i} .\) In particular, \(t<u_{i}\) for every \(i \geq m .\) Let \(n_{1}\) be the smallest integer for which \(n_{1} \geq m\) and \(x_{n_{1}}>t .\) For \(k=2,3,4, \ldots,\) let \(n_{k}\) be the smallest integer for which \(n_{k}>n_{k-1}\) and \(x_{n_{k}}>t .\) Then \(\left\{x_{n_{k}}\right\}_{k=1}^{\infty}\) is a subsequence of \(\left\{x_{i}\right\}_{i=m}^{\infty}\) which has a subsequential limit \(\lambda \geq t\). Since \(\lambda\) is also then a subsequential limit of \(\left\{x_{i}\right\}_{i=m}^{\infty},\) we have \(\lambda \in \Lambda\) and \(\lambda \geq t>s,\) contradicting \(s=\sup \Lambda .\) Hence we must have \(\lim \sup _{i \rightarrow \infty} x_{i}=\sup \Lambda .\) \(\quad\) Q.E.D.

    Exercise \(\PageIndex{27}\)

    Let \(\Lambda\) be the set of subsequential limits of a sequence \(\left\{x_{i}\right\}_{i=m}^{\infty} .\) Show that

    \[\liminf _{i \rightarrow \infty} x_{i}=\inf \Lambda .\]

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