7.3 Null Sequences

Sequences that converge to $0$ are simpler to work with than general sequences, and many of the convergence theorems for general sequences can be easily deduced from the properties of sequences that converge to $0$. In this section we will just consider sequences that converge to $0$.

7.11   Definition (Null sequence.) Let $f$ be a sequence in $\mbox{{\bf C}}$. We will say $f$ is a null sequence if and only if for every $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$ there is some $N\in\mbox{{\bf N}}$ such that for every $n\in\mbox{{\bf Z}}_{\geq N}$, $(\vert f(n)\vert<\varepsilon)$.

By comparing this definition with definition 7.10, you see that

$$(f\mbox{ is a null sequence })\iff(f\to 0).$$

Definition 7.11 is important. You should memorize it.

7.12   Definition (Dull sequence.) Let $f$ be a sequence in $\mbox{{\bf C}}$. We say $f$ is a dull sequence if and only if there is some $N\in\mbox{{\bf N}}$ such that for every $\varepsilon$ in $\mbox{${\mbox{{\bf R}}}^{+}$}$, and for every $n\in\mbox{{\bf Z}}_{\geq N}\;\; (\vert f(n)\vert<\varepsilon)$.

The definitions of null sequence and dull sequence use the same words, but they are not in the same order, and the definitions are not equivalent.

If $f$ satisfies condition (7.12), then whenever $n\geq N$,

$$\mbox{ for every }\varepsilon \mbox{ in } \mbox{${\mbox{{\bf R}}}^{+}$} \;\;\left(\vert f(n)\vert<\varepsilon\right).$$

If $\vert f(n)\vert\in\mbox{${\mbox{{\bf R}}}^{+}$}$, this condition would say $\vert f(n)\vert<\vert f(n)\vert$, which is false. Hence if $n\geq N$, then $\vert f(n)\vert\notin\mbox{${\mbox{{\bf R}}}^{+}$}$; i.e., if $n\geq N$, then $f(n)=0$. Hence a dull sequence has the property that there is some $N\in\mbox{{\bf N}}$ such that $f(n)=0$ for all $n\geq N$. Thus every dull sequence is a null sequence. The sequence

$$\left\{ 1,{1\over 2},{1\over 3},{1\over 4},0,0,0,0,0,\cdots\right\}$$

is a dull sequence, but

$$\left\{{1\over n}\right\}_{n\geq 1}=\left\{1,{1\over 2},{1\over 3},{1\over 4},{1\over 5},{1\over 6},{1\over 7},\cdots\right\}$$

is not a dull sequence. In the next theorem we show that $${\{{1\over n}\}_{n\geq 1}}$$ is a null sequence, so null sequences are not necessarily dull.

7.13   Theorem. . For all $a\in\mbox{{\bf C}}$, $${\left\{{a\over n}\right\}_{n\geq 1}}$$ is a null sequence

Proof: Let $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$. By the Archimedean property for $\mbox{{\bf R}}$, there is an $N\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ such that $${ N>{\vert a\vert \over \varepsilon}}$$. Then for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$,

$$n\geq N\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}{n} > ... ...ongrightarrow\hspace{1ex}$}{\vert a\vert\over n}<{\varepsilon},$$

so for all $${n\in\mbox{{\bf Z}}_{\geq N}\;\;\left( \left\vert{a\over n}\right\vert<\varepsilon\right)}$$. $\mid\!\mid\!\mid$

The difference between a null sequence and a dull sequence is that the `` $N$" in the definition of null sequence can (and usually does) depend on $\varepsilon$, while the ``$N$" in the definition of dull sequence depends only on $f$. To emphasize that $N$ depends on $\varepsilon$ (and also on $f$), I will often write $N(\varepsilon)$ or $N_f(\varepsilon)$ instead of $N$.

Here is another reformulation of the definition of null sequence.

7.14   Definition (Precision function.) Let $f$ be a complex sequence. Then $f$ is a null sequence if and only if there is a function $N_f\colon\mbox{${\mbox{{\bf R}}}^{+}$}\to\mbox{{\bf N}}$ such that

$$\mbox{ for all }\varepsilon > 0 \mbox{ and all } n\in\mbox{{\... ...Longrightarrow\hspace{1ex}$}\vert f(n)\vert<\varepsilon\right).$$

I will call such a function $N_f$ a precision function for $f$.

This formulation shows that in order to show that a sequence $f$ is a null sequence, you need to find a function $N_f\colon\mbox{${\mbox{{\bf R}}}^{+}$}\to\mbox{{\bf N}}$ such that

$$\mbox{ for all }n\in\mbox{{\bf N}}\; \left(n\geq N_f(\varepsi... ...Longrightarrow\hspace{1ex}$}\vert f(n)\vert<\varepsilon\right).$$

In the proof of theorem 7.13, for the sequence $${g\colon n\mapsto {a\over n}}$$ we had

$$N_g(\varepsilon)=\left(\mbox{ some integer } N\mbox{ such that } {\vert a\vert\over N}<\varepsilon\right).$$

This description for $N_g$ could be made more precise, but it is good enough for our purposes.

7.15   Theorem. If $\alpha\in\mbox{{\bf C}}\backslash\{0\}$, then the constant sequence $\tilde\alpha$ is not a null sequence.

Proof: If $\alpha\neq 0$, then ${{1\over 2}\vert\alpha\vert\in\mbox{${\mbox{{\bf R}}}^{+}$}}$. Suppose, to get a contradiction, that $\tilde\alpha$ is a null sequence. Then there is a number $N\in\mbox{{\bf N}}$ such that for all ${n\in\mbox{{\bf N}}\; \left(n\geq N\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\vert\tilde\alpha(n)\vert<{1\over 2}\vert\alpha\vert\right)}$. Then for all $n\in\mbox{{\bf N}}$,

$$\left(n\geq N\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$... ...$\hspace{1ex}\Longrightarrow\hspace{1ex}$}1<{1\over 2}\right).$$ (7.16)

If $n=N+1$ then (7.16) is false and this shows that $\tilde\alpha$ is not a null sequence. $\mid\!\mid\!\mid$

7.17   Theorem (Comparison theorem for null sequences.) Let $f,g$ be
complex sequences. Suppose that $f$ is a null sequence and that

$$\vert g(n)\vert\leq \vert f(n)\vert \mbox{ for all } n\in\mbox{{\bf N}}.$$

Then $g$ is a null sequence.

Proof: Since $f$ is a null sequence, there is a function $N_f\colon\mbox{${\mbox{{\bf R}}}^{+}$}\to\mbox{{\bf N}}$ such that for all $n\in\mbox{{\bf N}}$,

$$n\geq N_f(\varepsilon)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\vert f(n)\vert<\varepsilon.$$

Then

$$n\geq N_f(\varepsilon)\mbox{$\hspace{1ex}\Longrightarrow\hspa... ...e{1ex}\Longrightarrow\hspace{1ex}$}\vert g(n)\vert<\varepsilon.$$

Hence, we can let $N_g=N_f$. $\mid\!\mid\!\mid$

7.18   Example. We know that $n\leq 2^n$ for all $n\in\mbox{{\bf N}}$, so $${ {1\over {2^n}}\leq{1\over n}}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Since $${ \left\{ {1\over n}\right\}_{n\geq 1}}$$ is a null sequence, it follows from the comparison theorem that $${ \left\{ {1\over {2^n}}\right\}_{n\geq 1}}$$ is a null sequence. Also, since $${ {1\over {n^2+n}}\leq {1\over n}}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$, we see that $${\left\{ {1\over {n^2+n}}\right\}_{n\geq 1}}$$ is a null sequence.

7.19   Theorem (Root theorem for null sequences.) Let $f\colon\mbox{{\bf N}}\to[0,\infty)$ be a null sequence, and let $p\in\mbox{{\bf Z}}_{\geq 1}$. Then $${f^{{1\over p}}}$$ is a null sequence where $${f^{{1\over p}}(n)=\left(f(n)\right)^{{1\over p}}}$$ for all $n\in\mbox{{\bf N}}$.

Scratchwork: Let $${g=f^{{1\over p}}}$$. I want to find $N_g$ so that for all $n\in\mbox{{\bf N}}$ and all $\varepsilon \in\mbox{{\bf R}}^+$,

$$n\geq N_g(\varepsilon)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\vert g(n)\vert\leq\varepsilon$$

i.e.

$$n\geq N_g(\varepsilon) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\left\vert f^{{1\over p}}(n)\right\vert\leq\varepsilon$$

i.e.

$$n\geq N_g(\varepsilon) \mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}f(n)\leq\varepsilon^p.$$

This suggests that I should take $N_g(\varepsilon)=N_f(\varepsilon^p)$.

Proof: Let $f$ be a null sequence in $[0,\infty)$ and let $N_f$ be a precision function for $f$. Define $N_g\colon\mbox{${\mbox{{\bf R}}}^{+}$}\to\mbox{{\bf N}}$ by $N_g(\varepsilon)=N_f(\varepsilon^p)$ for all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$. Then for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} n\geq N_g(\varepsilon) &\mbox{$\Longrightarrow$}& n\geq N_f(\v... .../p}<\varepsilon \\ &\mbox{$\Longrightarrow$}& g(n)<\varepsilon. \end{eqnarray*}$$

Hence $N_g$ is a precision function for $g$. $\mid\!\mid\!\mid$

7.20   Examples. Let $c \in \mbox{{\bf R}}^+$. Then $${ \left\{{c^2\over n}\right\}_{n\geq 1}}$$ is a null sequence in $[0,\infty)$, so it follows that $${\left\{ {c\over {\sqrt n}}\right\}_{n\geq 1}}$$ is a null sequence.

Consider the sequence $${f\colon\mbox{{\bf Z}}_{\geq 1}\to\mbox{{\bf C}},\; f: n\mapsto n+{1\over 2}-\sqrt{n^2+n}}$$. For all $n\in\mbox{{\bf Z}}_{\geq 1}$,

$$\begin{eqnarray*} f(n)&=&\left(\left(n+{1\over 2}\right)-\sqrt{n^2+n}\right){{\l... ...qrt{n^2+n}}}={1\over {4\left(n+{1\over 2}+\sqrt{n^2+n}\right)}}. \end{eqnarray*}$$

Hence $${ \vert f(n)\vert\leq {1\over {4n}}\leq {1\over n}}$$, so it follows from the comparison theorem that $f$ is a null sequence.

Since $${ \left\{ {1\over {2^n}}\right\}_{n\geq 1}}$$ is a null sequence, it follows from the root theorem that $${\left\{\left( {1\over {\sqrt 2}}\right)^n\right\}_{n\geq 1}}$$ is a null sequence. Now $${.7^2=.49<{1\over 2}}$$, so $${.7<\sqrt{{1\over 2}}={1\over {\sqrt 2}}}$$ so

$${(.7)^n\leq\left( {1\over {\sqrt 2}}\right)^n}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$, and by another comparison test, $\{.7^n\}$ is a null sequence. Since $\Big((\vert\alpha\vert\leq .7)\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}(\vert\alpha^n\vert\leq .7^n)\Big)$, it follows that $\{\alpha^n\}_{n\geq 1}$ is a null sequence for all $\alpha\in\mbox{{\bf C}}$ with $\vert\alpha\vert\leq .7$.

You probably suspect that $\{\alpha^n\}$ is a null sequence for all $\alpha\in\mbox{{\bf C}}$ with $\vert\alpha\vert<1$. This is correct, but we will not prove it yet.

7.21   Exercise. A Show that the geometric sequences

$${ \left\{ \left( {{1+i}\over 2}\right)^n\right\}}$$ and
$${\left\{ \left( {{2+i}\over 3}\right)^n\right\}}$$ that are sketched above 7.1 are in fact null sequences.

7.22   Exercise. A Which, if any, of the sequences below are null sequences? Justify your answers.

a)

$\{ \sqrt{n+10000}-\sqrt n\}_{n\geq 1}$

b)

$${ \left\{ {{n^2+1}\over {n^3+3n}}\right\}_{n\geq 1}}$$

c)

$${ \left\{ {{n^2+6}\over {n^3+3n}}\right\}_{n\geq 1}}$$

7.23   Entertainment. Show that

$${\{(1-10^{-20})^n\}=\{.99999999999999999999^n\}}$$

is a null sequence. (If you succeed, you will probably find a proof that $\{\alpha^n\}$ is a null sequence whenever $\vert\alpha\vert<1$.) NOTE: If you use calculator operations, then $\{(1-10^{-20})^n\}$ is not a null sequence on most calculators.

It follows from remark 5.38 that we can add, subtract and multiply complex sequences, and that the usual associative, commutative, and distributive laws hold. If $f=\{f(n)\}$ and $g=\{g(n)\}$ then $f+g=\{f(n)+g(n)\}$ and $(fg)(n)=\{f(n)\cdot g(n)\}$. If $\alpha,\beta\in\mbox{{\bf C}}$ then the constant sequences $\tilde\alpha,\tilde\beta$ satisfy

$$\widetilde{\alpha+\beta}=\tilde\alpha+\tilde\beta,\quad\widetilde{\alpha\beta}=\tilde\alpha \tilde\beta.$$

7.24   Exercise. A Which of the field axioms are satisfied by addition and multiplication of sequences? Does the set of complex sequences form a field? (You know that the associative, distributive and commutative laws hold, so you just need to consider the remaining axioms.

7.25   Notation. If $f$ is a complex sequence, we define sequences $f^*$, $\mbox{\rm Re}f$, $\mbox{\rm Im}f$, and $\vert f\vert$ by

$$\begin{eqnarray*} f^*(n)&=&\left(f(n)\right)^* \mbox{ for all } n\in\mbox{{\bf N... ...f\vert(n)&=&\vert f(n)\vert \mbox{ for all } n\in\mbox{{\bf N}}. \end{eqnarray*}$$

7.26   Theorem. Let $f$ be a complex null sequence. Then $f^*$, $\mbox{\rm Re}f$, $\mbox{\rm Im}f$ and $\vert f\vert$ are all null sequences.

Proof: All four results follow by the comparison theorem. We have, for all $n\in\mbox{{\bf N}}$:

$$\begin{eqnarray*} \vert f^*(n)\vert&=&\vert(f(n))^*\vert = \vert f(n)\vert, \\ ... ...\vert(n)\right\vert&=&\vert f(n)\vert.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$