6.4 Properties of Limits.

In this section I will state some basic properties of limits. All of the statements listed here as assumptions are, in fact, theorems that can be proved from the definition of limits. I am omitting the proofs because of lack of time, and because the results are so plausible that you will probably believe them without a proof.

6.35   Definition (Constant sequence.) If $r$ is a real number then the sequence $\{r\}$ all of whose terms are equal to $r$ is called a constant sequence

$$\{r\}=\{r,r,r,\cdots\}.$$

It is an immediate consequence of the definition of convergence that

$$\{r\}\to r$$

for every real number $r$. (If $r_n=r$ for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ then $\vert r_n-r\vert=0<\epsilon$ for all $\epsilon$ in $\mbox{${\mbox{{\bf R}}}^{+}$}$ so $r_n$ approximates $r$ with an error smaller than $\epsilon$ for all $n\geq 1$. $\diamondsuit$.)

We have just proved

6.36   Theorem (Constant sequence rule.) If $\{r\}$ denotes a constant sequence of real numbers, then

$$\lim\{r\}=r.$$

6.37   Theorem (Null sequence rule.) Let $\alpha$ be a positive rational number. Then

$$\lim\left\{ {1\over {n^\alpha}}\right\}=0.$$

Proof: Let $\alpha$ be a positive rational number, and let $\epsilon$ be a generic positive number. By the monotonicity of powers , we hwve

$$\begin{eqnarray*} {1\over {n^\alpha}}<\epsilon &\hspace{1ex}\Longleftrightarrow\... ...>{1\over {\epsilon^{1\over \alpha}}}=\epsilon^{-{1\over\alpha}}. \end{eqnarray*}$$

By the Archimedian property for $\mbox{{\bf R}}$ there is an integer $N(\epsilon)$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

$$N(\epsilon)>\epsilon^{-{1\over\alpha}}.$$

Then for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$

$$n\geq N(\epsilon)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1e... ...ace{1ex}$}\left\vert {1\over {n^\alpha}}-0\right\vert<\epsilon.$$

Thus $${\lim\left\{ {1\over {n^\alpha}}\right\}=0}$$. $\diamondsuit$

6.38   Assumption (Sum rule for sequences.) Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences of real numbers. Then

$$\lim\{a_n +b_n\}=\lim\{a_n\}+\lim\{b_n\}$$

and

$$\lim\{a_n-b_n\}=\lim\{a_n\}-\lim\{b_n\}.$$

The sum rule is actually easy to prove, but I will not prove it. (You can probably supply a proof for it.)

Notice the hypothesis that $\{a_n\}$ and $\{b_n\}$ are convergent sequences. It is not true in general that

$$\lim\{a_n+b_n\}=\lim\{a_n\}+\lim\{b_n\}.$$

For example, the statement

$$\lim\{(-1)^n + (-1)^{n+1}\}=\lim\{(-1)^n\}+\lim\{(-1)^{n+1}\}$$

is false, since

$$\lim\{(-1)^n + (-1)^{n+1}\}=\lim\{0\}=0$$

but neither of the limits $\lim\{(-1)^n\}$ or $\lim\{(-1)^{n+1}\}$ exist.

6.39   Assumption (Product rule for sequences.) Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences. Then

$$\lim\{a_n\cdot b_n\}=\lim\{a_n\}\cdot\lim\{b_n\}.$$

An important special case of the product rule occurs when one of the sequences is constant: If $a$ is a real number, and $\{b_n\}$ is a convergent sequence, then

$$\lim\{ab_n\}=a\lim\{b_n\}.$$

The intuitive content of the product rule is that if $a_n$ approximates $a$ very well, and $b_n$ approximates $b$ very well, then $a_nb_n$ approximates $ab$ very well. It is somewhat tricky to prove this for a reason that is illustrated by the following example.

According to Maple,

$$\sqrt{99999999}=9999.99994999999987499\cdots$$

so $9999.9999$ approximates $\sqrt{99999999}$ with 4 decimal accuracy. Let

$$a=b=9999.9999,$$

and let

$$A=B=\sqrt{99999999}.$$

Then $a$ approximates $A$ with 4 decimal accuracy and $b$ approximates $B$ with 4 decimal accuracy. But

$$AB=99999999$$

and

$$ab=99999998.00000001$$

so $ab$ does not approximate $AB$ with an accuracy of even one decimal.

6.40   Assumption (Quotient rule for sequences.) Let $\{a_n\}$ and $\{b_n\}$ be convergent real sequences such that $b_n\neq 0$ for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ and $\lim\{b_n\}\neq 0$. Then

$$\lim\left\{ {{a_n}\over {b_n}}\right\}={{\lim\{a_n\}}\over {\lim\{b_n\}}}.$$

The hypotheses here are to be expected. If some term $b_n$ were zero, then $${ \left\{ {{a_n}\over {b_n}}\right\}}$$ would not be a sequence, and if $\lim\{b_n\}$ were zero, then $${ {{\lim\{a_n\}}\over {\lim\{b_n\}}}}$$ would not be defined.

6.41   Assumption (Inequality rule for sequences.) Let $\{a_n\}$ and $\{b_n\}$ be convergent sequences. Suppose there is an integer $N$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

$$a_n\leq b_n \mbox{ for all } n \mbox{ in } \mbox{{\bf Z}}_{\geq N}.$$

Then

$$\lim\{a_n\}\leq\lim\{b_n\}.$$

The most common use of this rule is in situations where

$$0\leq b_n \mbox{ for all } n$$

and we conclude that

$$0\leq\lim\{b_n\}.$$

6.42   Assumption (Squeezing rule for sequences.) Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be three real sequences. Suppose there is an integer $N$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

$$a_n \leq b_n \leq c_n \mbox{ for all }n \in \mbox{{\bf Z}}_{\geq N}.$$ (6.43)

Suppose further, that $\{a_n\}$ and $\{c_n\}$ both converge to the same limit $L$. Then $\{b_n\}$ also converges to $L$.

If we knew that the middle sequence, $\{b_n\}$ in the squeezing rule was convergent, then we would be able to prove the squeezing rule from the inequality rule, since if all three sequences $\{a_n\}$, $\{b_n\}$ and $\{c_n\}$ converge, then it follows from (6.43) that

$$\lim \{a_n\} \leq \lim \{b_n\} \leq \lim \{c_n\},$$

i.e.

$$L \leq \lim \{b_n\} \leq L$$

and hence $\lim\{b_n\} = L$. The power of the squeezing rule is that it allows us to conclude that a limit exists.

6.44   Definition (Translate of a sequence.) Let $\{a_n\}$ be a real sequence, and let $p\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. The sequence $\{a_{n+p}\}$ is called a translate of $\{a_n\}$.

6.45   Example. If

$$\{a_n\}=\left\{{1\over {n^2}}\right\}=\left\{1,{1\over 4},{1\over 9},{1\over {16}},{1\over {25}},\cdots\right\}$$

then

$$\{a_{n+2}\}=\left\{ {1\over {(n+2)^2}}\right\}=\left\{{1\over 9},{1\over {16}},{1\over {25}},\cdots\right\}.$$

If

$$\{b_n\}=\{(-1)^n\}$$

then

$$\{b_{n+2}\}=\{(-1)^{n+2}\}=\{(-1)^n\}=\{b_n\}.$$

6.46   Theorem (Translation rule for sequences.) Let $\{a_n\}$ be a convergent sequence of real numbers, and let $p$ be a positive integer. Then $\{a_{n+p}\}$ is convergent and

$$\lim\{a_{n+p}\}=\lim\{a_n\}.$$

Proof: Suppose $\lim\{a_n\}=a$, and let $\epsilon$ be a generic element in $\mbox{${\mbox{{\bf R}}}^{+}$}$. Then we can find an integer $N(\epsilon)$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

$$\vert a_n-a\vert<\epsilon \mbox{ for all } n \mbox{ in } \mbox{${\mbox{{\bf Z}}}^{+}$}\mbox{ with } n\geq N(\epsilon).$$

If $n\geq N(\epsilon)$ then $n+p\geq N(\epsilon)+p\geq N(\epsilon)$ so

$$\vert a_{n+p}-a\vert<\epsilon.$$

This shows that $\lim\{a_{n+p}\}=a=\lim\{a_n\}.$ $\diamondsuit$

6.47   Example. The sequence

$$\{a_n\}=\left\{{1\over {n+4}}\right\}=\left\{ {1\over 5},{1\over 6},{1\over 7},\cdots\right\}$$

is a translate of the sequence $${\left\{{1\over n}\right\}}$$. Since $${ \lim\left\{ {1\over n} \right\}=0}$$ it follows from the translation theorem that $${\lim\left\{{1\over {n+4}}\right\}=0}$$ also.

6.48   Theorem ($n$th root rule for sequences.) Let $a$ be a positive number then

$$\lim\left\{a^{1\over n}\right\}=1.$$

Proof: Case 1: Suppose $a=1$. Then

$$\lim\left\{a^{1\over n}\right\}=\lim\{1\}=1.$$

Case 2: Suppose $a > 1$, so that $a^{1\over n} > 1$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. Let $\epsilon$ be a generic positive number, and let $n$ be a generic element of $\mbox{${\mbox{{\bf Z}}}^{+}$}$. Since $\ln$ is strictly increasing on $\mbox{${\mbox{{\bf R}}}^{+}$}$ we have

$$\left(a^{1\over n}-1<\epsilon\right) \mbox{$\Longleftrightarrow$} \left(a^{1\over n}<1+\epsilon\right) \hspace{1ex}\Longleftrightarrow\hspace{1ex}\left(\ln(a^{1\over n})<\ln(1+\epsilon)\right) \mbox{{}}$$

$$\mbox{$\Longleftrightarrow$} {1\over n}\ln(a)<\ln(1+\epsilon)\mbox{{}}$$

$$\mbox{$\Longleftrightarrow$} {\ln(a) \over \ln(1+\epsilon)} < n.$$ (6.49)

(In the last step I used the fact that $\ln(1+\epsilon)>0$ if $\epsilon >0$.) By the Archimedean property for $\mbox{{\bf R}}$ there is an integer $N(\epsilon)$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

$${\ln(a) \over \ln(1+\epsilon)} < N(\epsilon).$$

For all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ we have

$$\begin{eqnarray*} n \geq N(\epsilon)& \mbox{$\Longrightarrow$}& {\ln(a) \over \... ...pace{1ex}$} \left\vert a^{1\over n} - 1 \right\vert < \epsilon. \end{eqnarray*}$$

Hence $${\lim\left\{a^{1\over n}\right\}=1}$$.

Case 3: Suppose $0<a<1$. Then $a^{-1}>1$ so by Case 2, we have

$$\begin{eqnarray*} \lim\left\{ a^{1\over n}\right\} &=& \lim\left\{ {1\over {(a^{... ...}}\over {\lim\left\{(a^{-1})^{1\over n}\right\}}}={1\over 1}=1. \end{eqnarray*}$$

Thus, in all cases, we have

$$\lim\left\{ a^{1\over n}\right\}=1.\mbox{ $\diamondsuit$}$$