6.3 Convergence of Sequences

6.19   Definition ($\{a_n\}$ converges to $L$.)

Let $\{a_n\}$ be a sequence of real numbers, and let $L$ be a real number. We say that $\{a_n\}$ converges to $L$ if for every positive number $\epsilon$ there is a number $N(\epsilon)$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that all of the numbers $a_n$ for which $n\geq N(\epsilon)$ approximate $L$ with an error smaller than $\epsilon$. We denote the fact that $\{a_n\}$ converges to $L$ by the notation

$$\{a_n\}\to L.$$

Thus ``$\{a_n\}\to L$" means:

For every $\epsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$ there is a number $N(\epsilon)$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

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

Since

$$\vert a_n-L\vert =\vert(a_n-L)-0\vert=\Big\vert \vert a_n-L\vert -0 \Big\vert,$$

it follows immediately from the definition of convergence that

$$(\{a_n\}\to L)\hspace{1ex}\Longleftrightarrow\hspace{1ex}(\{a... ...ace{1ex}\Longleftrightarrow\hspace{1ex}(\vert a_n-L\vert\to 0).$$

We will make frequent use of these equivalences.

6.20   Example. If $a \in \mbox{${\mbox{{\bf R}}}^{+}$}$ then

$$\left\{ a^3\left( 1+{1\over n}\right)\left(1+{1\over {2n}}\right)\right\}\to a^3.$$

Proof: Let $\epsilon$ be a generic element of $\mbox{${\mbox{{\bf R}}}^{+}$}$. I must find a number $N(\epsilon)$ such that

$$\Big\vert a^3\left(1+{1\over n}\right)\left(1+{1\over {2n}}\right)-a^3\Big\vert <\epsilon$$ (6.21)

whenever $n\geq N(\epsilon)$. Well, for all $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$

$$\Big\vert a^3\left(1+{1\over n}\right)\left(1+{1\over {2n}}\right)-a^3\Big\vert = \Big\vert a^3\left(1+{3\over {2n}}+{1\over {2n^2}}\right)-a^3\Big\vert\mbox{{}}$$

$$= \Big\vert a^3 \left({3\over {2n}}+{1\over {2n^2}}\right)\Big\vert=a^3\left( {3\over {2n}} +{1\over {2n^2}}\right)\mbox{{}}$$

$$\leq a^3\left({3\over{2n}}+{1\over {2n}}\right)={{2a^3}\over n}.$$ (6.22)

Now for every $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ we have

$$\left( {{2a^3}\over n}<\epsilon\right)\hspace{1ex}\Longleftrightarrow\hspace{1ex}\left( {{2a^3}\over \epsilon}<n\right),$$

and by the Archimedean property of $\mbox{{\bf R}}$ there is some integer $N(\epsilon)$ such that $${ {{2a^3}\over \epsilon}<N(\epsilon)}$$. For all $n\geq N(\epsilon)$ we have

$$\left(n\geq N(\epsilon)\right)\mbox{$\hspace{1ex}\Longrightar... ...space{1ex}$}\left(\left({{2a^3}\over n}\right)<\epsilon\right),$$

so by (6.22)

$$\left(n\geq N(\epsilon)\right)\mbox{$\hspace{1ex}\Longrightar... ...{1\over 2n}\right) -a^3\Big\vert \leq {{2a^3}\over n}<\epsilon.$$

Hence by the definition of convergence we have

$$\left\{a^3\left(1+{1\over n}\right)\left(1+{1\over {2n}}\right)\right\}\to a^3.\mbox{ $\diamondsuit$}$$ (6.23)

A very similar argument can be used to show that

$$\left\{a^3\left(1-{1\over n}\right)\left(1-{1\over {2n}}\right)\right\}\to a^3.$$ (6.24)

6.25   Example. In the eighteenth century the rather complicated argument just given would have been stated as

If $n$ is infinitely large, then $${a^3\left( 1 + {1\over n} \right) \left( 1 + {1\over 2n} \right) = a^3}$$.

The first calculus text book (written by Guillaume François de l'Hôpital and published in 1696) sets forth the postulate

Grant that two quantities, whose difference is an infinitely small quantity, may be taken (or used) indifferently for each other: or (which is the same thing) that a quantity which is increased or decreased only by an infinitely small quantity, may be considered as remaining the same[35, page 314].

If $n$ is infinite, then $${1 \over n}$$ is infinitely small, so $${\Big(1 + {1\over n } \Big) = 1}$$, and similarly $${\Big(1 + {1\over 2n} \Big) = 1}$$. Hence

$$a^3\left( 1 + {1\over n} \right) \left( 1 + {1\over 2n} \right) = a^3\cdot 1\cdot 1 = a^3.$$

There were numerous objections to this sort of reasoning. Even though $${\Big(1 + {1\over n } \Big) = 1}$$, we do not have $${\Big(1 + {1\over n}\Big) - 1 = 0}$$, since

$${\Big( 1+{1\over n}\Big) - 1 \over {1\over n}} = 1.$$

It took many mathematicians working over hundreds of years to come up with our definition of convergence.

6.26   Theorem (Uniqueness theorem for convergence.) Let $\{a_n\}$ be a sequence of real numbers, and let $a,b$ be real numbers. Suppose

$$\{a_n\}\to a \mbox{ and } \{a_n\}\to b.$$

Then $a=b$.

Proof: Suppose $\{a_n\}\to a$ and $\{a_n\}\to b$. By the triangle inequality

$$\vert a-b\vert=\vert(a-a_n)-(b-a_n)\vert\leq \vert a-a_n\vert+\vert b-a_n\vert.$$ (6.27)

Let $\epsilon$ be a generic element of $\mbox{${\mbox{{\bf R}}}^{+}$}$. Then ${{\epsilon\over 2}}$ is also in $\mbox{${\mbox{{\bf R}}}^{+}$}$. Since $\{a_n\}\to a$, there is a number $N({\epsilon\over 2})$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

$$\vert a-a_n\vert <{\epsilon\over 2} \mbox{ for all } n\geq\textstyle{N({\epsilon\over 2})}.$$ (6.28)

Since $\{a_n\}\to b$ there is a number ${M({\epsilon\over 2})}$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that

$$\vert b-a_n\vert<{\epsilon\over 2} \mbox{ for all } n\geq{\textstyle M({\epsilon\over 2})}.$$ (6.29)

Let $P(\epsilon)$ be the larger of $N({\epsilon\over 2})$ and $M({\epsilon\over 2})$. If $n$ is a positive integer and $n\geq P(\epsilon)$ then by (6.27), (6.28), and (6.29), we have

$$\vert a-b\vert \leq \vert a-a_n\vert+\vert b-a_n\vert<{\epsilon\over 2}+{\epsilon\over 2}=\epsilon.$$

Since this holds for all $\epsilon$ in $\mbox{${\mbox{{\bf R}}}^{+}$}$, we have $a=b$. $\diamondsuit$

6.30   Definition (Limit of a sequence.) Let $\{a_n\}$ be a sequence of real numbers. If there is a number $a$ such that $\{a_n\}\to a$, we write $\lim\{a_n\}=a$. The uniqueness theorem for convergence shows that this definition makes sense. If $\lim\{a_n\}=a$, we say $a$ is the limit of the sequence $\{a_n\}$.

6.31   Definition (Convergent and divergent sequence.) Let $\{a_n\}$ be a sequence of real numbers. If there is a number $a$ such that $\{a_n\}\to a$, we say that $\{a_n\}$ is a convergent sequence. If there is no such number $a$, we say that $\{a_n\}$ is a divergent sequence.

6.32   Example. It follows from example 6.20 that

$$\lim\left\{ a^3\left(1+{1\over n}\right)\left(1+{1\over {2n}}\right)\right\}=a^3$$

for all $a$ in $\mbox{${\mbox{{\bf R}}}^{+}$}$. Hence $${\left\{ a^3\left(1+{1\over n}\right)\left(1+{1\over {2n}}\right)\right\}}$$ is a convergent sequence for each $a$ in $\mbox{${\mbox{{\bf R}}}^{+}$}$.

The sequence $\{n\}$ is a divergent sequence. To see this, suppose there were a number $a$ such that $\{n\}\to a$.

Then we can find a number ${N({1\over 3})}$ such that

$$\vert n-a\vert<{1\over 3} \mbox{ for all } n\geq {\textstyle N({1\over 3})}.$$

In particular

$$\left\vert\textstyle{N({1\over 3})}-a\right\vert <{1\over 3} ... ...eft({\textstyle N({1\over 3})}+1\right)-a\right\vert<{1\over 3}$$

(since ${N({1\over 3})+1}$ is an integer greater than ${N({1\over 3})}$). Hence, by the triangle inequality

$$\begin{eqnarray*} 1&=&\vert 1\vert = \left\vert \left({\textstyle N({1\over 3})}... ...e N({1\over 3})} -a\right\vert<{1\over 3}+{1\over 3}={2\over 3} \end{eqnarray*}$$

i.e., $${1<{2\over 3}}$$ which is false.

Since the assumption $\{n\}\to a$ has led to a contradiction, it is false that $\{n\}\to a$. $\diamondsuit$

6.33   Exercise. A Let $\{a_n\}$ be a sequence of real numbers, and let $a$ be a real number. Suppose that as $n$ gets larger and larger, $a_n$ gets nearer and nearer to $a$, i.e., suppose that for all $m$ and $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$

$$(n>m)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(\vert a_n-a\vert<\vert a_m-a\vert).$$

Does it follow that $\{a_n\}$ converges to $a$?

6.34   Exercise. For each of the sequences below, calculate the first few terms, and make a guess as to whether or not the sequence converges. In some cases you will need to use a calculator. Try to explain the basis for your guess. (If you can prove your guess is correct, do so, but in several cases the proofs involve more mathematical knowledge than you now have.)

$${\{a_n\} = \{(-1)^n\}.}$$

$${\{c_n\} = \left\{1-{1\over 2}+{1\over 3}-{1\over 4}+\cdots +{{(-1)^{n+1}}\over n}\right\}.}$$

$${\{d_n\} =\left\{1+{1\over {2^2}}+{1\over {3^2}}+\cdots +{1\over {n^2}}\right\} }$$
This problem was solved by Leonard Euler (1707-1783)[18, pp138-139].

$${\{e_n\} = \left\{1+{1\over 3}+{1\over {3^2}}+\cdots +{1\over {3^{n-1}}}\right\}.}$$

$${\{f_n\} = \left\{\left(1+{1\over n}\right)^n\right\} }$$
This problem was solved by Jacob Bernoulli (1654-1705)[8, pp94-97].

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