7.5 Theorems About Convergent Sequences

7.29   Remark. Let $f$ be a complex sequence, and let $L\in\mbox{{\bf C}}$. Then the following three statements are equivalent.

a)

$f\to L$

b)

$f-\tilde L$ is a null sequence.

c)

$\vert f-\tilde L\vert$ is a null sequence.

Proof: By definition 7.10, `` $f\to L$" means

for every $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$ there is some $N\in\mbox{{\bf N}}$ such that
for every $n\in\mbox{{\bf Z}}_{\geq N}$, $\left(\vert f(n)-L\vert<r\right)$.

By definition 7.11, `` $f-\tilde L$ is a null sequence" means

$$\; \mbox{ for every } \varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}\mbox{ there is some } N\in\mbox{{\bf N}}\mbox{ such that }$$

$$\hspace{.5in} \mbox{ for every } \;n\in\mbox{{\bf Z}}_{\geq N}, \vert(f-\tilde L)(n)\vert<\varepsilon.$$ (7.30)

Both definitions say the same thing. If we write out the definition for `` $\vert f-\tilde L\vert$ is a null sequence" we get (7.30) with `` $\vert(f-\tilde L)(n)\vert<\varepsilon$" replaced by
`` $\left\vert \vert f-\tilde L\vert(n)\right\vert<\varepsilon$." Since

$$\vert(f-\tilde L)(n)\vert=\vert f(n)-L\vert=\left\vert \vert f-\tilde L\vert(n)\right\vert,$$

conditions b) and c) are equivalent. $\mid\!\mid\!\mid$

7.31   Theorem (Decomposition theorem.) Let $f$ be a convergent complex sequence. Then we can write

$$f=k+\tilde c$$

where $k$ is a null sequence, and $\tilde c$ is a constant sequence. If $f\to L$, then $c= L$.

Proof: $f=(f-\tilde L)+\tilde L$. $\mid\!\mid\!\mid$

7.32   Theorem (Sum theorems for convergent sequences.) Let $\alpha\in\mbox{{\bf C}}$ and let $f,g$ be convergent complex sequences. Say $f\to L$ and $g\to M$. Then $f+g$, $f-g$ and $\alpha f$ are convergent and

$$\begin{eqnarray*} f+g&\to &L+M \\ f-g&\to& L-M \\ \alpha f&\to&\alpha L. \end{eqnarray*}$$

Proof: Suppose $f\to L$ and $g\to M$. By the decomposition theorem, we can write

$$f = k + \tilde{L} \mbox{ and }g = p + \tilde{M}$$

where $k$ and $p$ are null sequences. Then

$$(f\pm g) - (\widetilde{L \pm M}) = (k + \tilde{L})\pm(p+\tilde{M}) -( \tilde{L} \pm \tilde{M}) = k \pm p.$$

By the sum theorem for null sequences, $k\pm p$ is a null sequence, so $(f\pm g) -\widetilde{L\pm M}$ is a null sequence, and hence $f \pm g \to L\pm M$. $\mid\!\mid\!\mid$

7.33   Exercise. Prove the last statement in theorem 7.32; i.e., show that if $f\to L$ then $\alpha f\to\alpha L$ for all $\alpha\in\mbox{{\bf C}}$.

7.34   Theorem (Product theorem for convergent sequences.) Let $f,g$ be convergent complex sequences. Suppose $f\to L$ and $g\to M$. Then $fg$ is convergent and $fg\to LM$.

Proof: Suppose $f\to L$ and $g\to M$. Write $f=k+\tilde L$, $g=p+\tilde M$ where $k,p$ are null sequences. Then

$$\begin{eqnarray*} fg&=&(k+\tilde L)(p+\tilde M) \\ &=&kp+\tilde Lp+\tilde Mk+\tilde L\tilde M \\ &=&kp+Lp+Mk+\widetilde{LM}. \end{eqnarray*}$$

Now $kp$, $Lp$ and $Mk$ are null sequences by the product theorem and sum theorem for null sequences, and $\widetilde{LM}\to LM$, so by several applications of the sum theorem for convergent sequences,

$$fg\to 0+0+0+LM; \mbox{ i.e. } fg\to LM.\mbox{ $\mid\!\mid\!\mid$}$$

7.35   Theorem (Uniqueness theorem for convergent sequences.) Let
$f$ be a complex sequence, and let $L,M\in\mbox{{\bf C}}$. If $f\to L$ and $f\to M$, then $L=M$.

Proof: Suppose $f\to L$ and $f\to M$. Then $f-\tilde L$ and $f-\tilde M$ are null sequences, so $(f-\tilde L)-(f-\tilde M)=\tilde M-\tilde L=\widetilde{M-L}$ is a null sequence. Hence, by theorem 7.15, $M-L=0$; i.e., $L=M$. $\mid\!\mid\!\mid$

7.36   Definition (Limit of a sequence.) Let $f$ be a convergent sequence. Then the unique complex number $L$ such that $f\to L$ is denoted by $\lim f$ or $\lim\{f(n)\}$.

7.37   Remark. It follows from the sum and product theorems that if $f$ and $g$ are convergent sequences, then

$$\lim(f\pm g)=\lim f\pm\lim g$$

and

$$\lim(f\cdot g)=\lim f\cdot\lim g$$

and

$$\lim cf=c\lim f.$$

7.38   Warning. We have only defined $\lim f$ when $f$ is a convergent sequence. Hence $\lim\{i^n\}$ is ungrammatical and should not be written down. (We showed in theorem 7.7 that $\{i^n\}$ diverges.) However, it is a standard usage to say ``$\lim f$ does not exist'' or ``$\lim\{f(n)\}$ does not exist'' to mean that the sequence $f$ has no limit. Hence we may say ``$\lim\{i^n\}$ does not exist''.

7.39   Theorem. Let $f$ be a complex sequence. Then $f$ is convergent if and only if both $\mbox{\rm Re}f$ and $\mbox{\rm Im}f$ are convergent. Moreover,

$$\lim f = \lim\mbox{\rm Re}f+i\lim\mbox{\rm Im}f,$$ (7.40)

$$\lim\mbox{\rm Re}f = \mbox{\rm Re}(\lim f),$$

$$\lim\mbox{\rm Im}f = \mbox{\rm Im}(\lim f).$$

Proof: If $\mbox{\rm Re}f$ and $\mbox{\rm Im}f$ are convergent, then it follows from the sum theorem for convergent sequences that $f$ is convergent and (7.40) is valid.

Suppose that $f\to L$. Then $f-\tilde L$ is a null sequence, so $\mbox{\rm Re}(f-\tilde L)$ is a null sequence (by Theorem 7.26). For all $n\in\mbox{{\bf N}}$,

$$\mbox{\rm Re}(f-\tilde L)(n)=\mbox{\rm Re}\left(f(n)-L\right)... ...-\mbox{\rm Re}L=(\mbox{\rm Re} f-\widetilde{\mbox{\rm Re}L})(n)$$

so $(\mbox{\rm Re}f-\widetilde{\mbox{\rm Re}L})=\mbox{\rm Re}(f-\tilde L)$ is a null sequence and it follows that $\mbox{\rm Re}f$ converges to $\mbox{\rm Re}L$. A similar argument shows that $\mbox{\rm Im} f\to\mbox{\rm Im}L$. $\mid\!\mid\!\mid$

7.41   Definition (Bounded sequence.) A sequence $f$ in $\mbox{{\bf C}}$ is bounded, if there is a disc $\overline{D}(0,B)$ such that $f(n)\in\overline D(0,B)$ for all $n\in\mbox{{\bf N}}$; i.e., $f$ is bounded if there is a number $B\in[0,\infty)$ such that

$$\vert f(n)\vert\leq B \mbox{ for all } n\in\mbox{{\bf N}}.$$ (7.42)

Any number $B$ satisfying condition (7.42) is called a bound for $f$.

7.43   Examples. $${\left\{ {{i^nn}\over {n+1}}\right\}}$$ is bounded since $${\left\vert{{i^nn}\over {n+1}}\right\vert={n\over {n+1}}\leq 1}$$ for all $n\in\mbox{{\bf N}}$. The sequence $\{n\}$ is not bounded since the statement $\vert n\vert\leq B$ for all $n\in\mbox{{\bf N}}$ contradicts the Archimedean property of $\mbox{{\bf R}}$. Every constant sequence $\{\tilde L\}$ is bounded. In fact, $\vert L\vert$ is a bound for $\tilde L$.

7.44   Exercise (Null-times-bounded theorem.) A Show that if $f$ is a null sequence in $\mbox{{\bf C}}$, and $g$ is a bounded sequence in $\mbox{{\bf C}}$ then $fg$ is a null sequence.

The next theorem I want to prove is a quotient theorem for convergent sequences. To prove this, I will need some technical results.

7.45   Theorem (Reverse triangle inequality.) Let $\alpha,\beta\in\mbox{{\bf C}}$, then

$$\vert\alpha-\beta\vert\geq\vert\alpha\vert-\vert\beta\vert.$$

Proof: By the triangle inequality.

$$\vert\alpha\vert=\vert(\alpha-\beta)+\beta\vert\leq\vert\alpha-\beta\vert+\vert\beta\vert.$$

Hence,

$$\vert\alpha\vert-\vert\beta\vert\leq\vert\alpha-\beta\vert.\mbox{ $\mid\!\mid\!\mid$}$$

7.46   Lemma. Let $f$ be a convergent sequence that is not a null sequence; i.e., $f\to L$ where $L\neq 0$. Suppose $f(n)\neq 0$ for all $n\in\mbox{{\bf N}}$. Then $${{1\over f}}$$ is a bounded sequence.

Proof: Since $f\to L$, we know that $f-\tilde L$ is a null sequence. Let $N_{f-\tilde L}$ be a precision function for $f-\tilde L$. Then for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} n\geq N_{f-\tilde L}\left( {{\vert L\vert}\over 2}\right) &\mb... ...eft\vert {1\over {f(n)}}\right\vert\leq {2\over {\vert L\vert}}; \end{eqnarray*}$$

i.e., if $${M=N_{f-\tilde L}\left({{\vert L\vert}\over 2}\right)}$$, then

$$n\geq M\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}\left\vert {1\over {f(n)}}\right\vert\leq {2\over {\vert L\vert}}.$$

Let

$$B=\max\left( {2\over {\vert L\vert}}, \max_{0\leq m\leq M}\left\vert{1\over {f(m)}}\right\vert\right).$$

Then $${\left\vert {1\over {f(m)}}\right\vert\leq B}$$ for $m\in\mbox{{\bf Z}}_{0\leq m\leq M}$ and $${\left\vert {1\over {f(m)}}\right\vert\leq B}$$ for $m\in\mbox{{\bf Z}}_{\geq M}$, so
$\left\vert {1\over {f(m)}}\right\vert\leq B$ for all $m\in\mbox{{\bf Z}}_{\geq 0}=\mbox{{\bf N}}$, and hence $${{1\over f}}$$ is bounded. $\mid\!\mid\!\mid$

7.47   Theorem (Reciprocal theorem for convergent sequences.) Let $g$ be a complex sequence. Suppose that $g\to L$ where $L\neq 0$, and that $g(n)\neq 0$ for all $n\in\mbox{{\bf N}}$. Then $${{1\over g}}$$ is convergent, and $${ {1\over g}\to {1\over L}}$$.

Proof: By the preceding lemma, $${{1\over g}}$$ is a bounded sequence, and since $g\to L$, we know that $g-\tilde L$ is a null sequence. Hence $${(g-\tilde L)\cdot{1\over g}=\tilde 1-{L\over g}}$$ is a null sequence, and it follows that $${{L\over g}\to 1}$$. Then we have

$${1\over g}={1\over L}\cdot {L\over g}\to {1\over L}\cdot 1={1\over L};$$

i.e., $${ {1\over g}\to {1\over L}}$$. $\mid\!\mid\!\mid$

7.48   Exercise (Quotient theorem for convergent sequences.) The following statement isn't quite true. Supply the missing hypotheses and prove the corrected statement.

Let $f,g$ be convergent complex sequences. If $f\to L$ and $g\to M$, then $${{f\over g}}$$ is convergent and $${ {f\over g}\to {L\over M}}$$.

7.49   Exercise. A

a)

Let $f,g$ be complex sequences. Show that if $f$ converges and $g$ diverges, then $f+g$ diverges.

b)

Show that if $f$ converges and $g$ diverges, then $fg$ does not necessarily diverge.

7.50   Exercise. A Let $f$ be a divergent complex sequence. Show that if $c\in\mbox{{\bf C}}\backslash\{0\}$, then $cf$ is divergent.

7.51   Example. Let $f\colon\mbox{{\bf Z}}_{\geq 1}\to\mbox{{\bf C}}$ be defined by

$${f(n)={{n^2+in+1}\over {3n^2+2in-1}}}.$$ (7.52)

Then

$$f(n)={{n^2\left(1+{i\over n}+{1\over {n^2}}\right)}\over {n^... ...r {n^2}}}\over {\left(3-{1\over {n^2}}\right)+{{2i}\over n}}}.$$ (7.53)

Hence $f$ can be written as a quotient of two sequences:

$$h\colon n\mapsto 1+{i\over n}+{1\over {n^2}}$$

and

$${g\colon n\mapsto\left(3-{1\over {n^2}}\right)+{{2i}\over n}}$$

where $g(n)\neq 0$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Since

$${h=\tilde 1+i\left\{ {1\over n}\right\}_{n\geq 1}+\left\{{1\over n}\right\}_{n\geq 1}\cdot\left\{ {1\over n}\right\}_{n\geq 1}}$$

and

$$g=\tilde 3-\left\{{1\over n}\right\}_{n\geq 1}+2i\left\{{1\over n}\right\}_{n\geq 1},$$

it follows from numerous applications of product and sum rules that $h\to 1$ and $g\to 3\neq 0$ and hence $${f={h\over g}\to{1\over 3}}$$. Once I have expressed $f(n)$ in the final form in (7.53), I can see what the final result is, and I will usually just write

$$\left\{f(n)\right\}=\left\{ {{1+{i\over n}+{1\over {n^2}}}\ov... ...+{{2i}\over n}}}\right\} \to {{1+0+0}\over {3-0+0}}={1\over 3}.$$

7.54   Example. Let $g\colon\mbox{{\bf N}}\to\mbox{{\bf C}}$ be the sequence

$$g=\left\{{{2^n+4^n}\over {4^n+6^n}}\right\}.$$ (7.55)

Then for all $n\in\mbox{{\bf N}}$,

$$g(n)={{2^n+4^n}\over {4^n+6^n}}={{4^n\left( {{2^n}\over {4^n}... ...ver 2}\right)^n+1}\over {\left( {2\over 3}\right)^n+1}}\right).$$

Since $${\left\vert{2\over 3}\right\vert<.7}$$, I know $${\left\{\left({2\over 3}\right)^n\right\}\to 0}$$ and $${\left\{\left({1\over 2}\right)^n\right\}\to 0}$$ so

$$\{g(n)\}=\left\{\left( {2\over 3}\right)^n {{\left(\left( {... ...}\right)^n+1\right)}}\right\}\to 0 \cdot {{0+1}\over {0+1}}=0.$$

In the last two examples, I was motivated by the following considerations. I think: In the numerator and denominator for (7.52), for large $n$ the `` $n^2$" term overwhelms the other terms - so that's the term I factored out. In the numerator of (7.55), the overwhelming term is $4^n$, and in the denominator, the overwhelming term is $6^n$ so those are the terms I factored out.

7.56   Exercise. A Let $\{f(n)\}$ be a sequence of non-negative numbers and suppose $\{f(n)\}\to L$ where $L>0$. Prove that $${\{\sqrt{f(n)}\}\to\sqrt L}$$. (NOTE: The case $L=0$ follows from the root theorem for null sequences.

7.57   Exercise. A Investigate the sequences below, and find their limits if they have any.

a)

$${f=\left\{ {{1+3n+3in^2}\over {1+2in+5n^2}}\right\}_{n\geq 1}}$$

b)

$${g=\left\{ {{n^2+3in+1}\over {n^3+n+i}}\right\}_{n\geq 1}}$$

c)

$${h=\left\{ {{\left(4+{1\over n}\right)^2-16}\over {\left(3+{i\over n}\right)^2-9}}\right\}_{n\geq 1}}$$

d)

$${k=\left\{ \sqrt{1+{1\over n}}\right\}_{n\geq 1}}$$

e)

$${l=\left\{\sqrt{n^2+n}-n\right\}_{n\geq 1}}$$

7.58   Exercise. A Show that the sum of two bounded sequences is a bounded sequence.

7.59   Theorem (Convergent sequences are bounded.) Let $\{\alpha_n\}$ be a convergent complex sequence. Then $\{\alpha_n\}$ is bounded.

Proof: I will show that null sequences are bounded and leave the general case to you. Let $f$ be a null sequence and let $N_f$ be a precision function for $f$.

Let

$$B=\max\left( 1,\;\max_{0\leq j\leq N_f(1)}\left(\vert f(j)\vert\right)\right).$$

I claim that $B$ is a bound for $f$. If $n\in\mbox{{\bf Z}}_{0\leq j\leq N_f(1)}$, then

$${\vert f(n)\vert\leq\max_{0\leq j\leq N_f(1)}\left(\vert f(j)\vert\right)\leq B}.$$

If $n\in\mbox{{\bf Z}}_{\geq N_f(1)}$, then $n \geq N_f(1)$, so $\vert f(n)\vert\leq 1\leq B$. Hence

$$\vert f(n)\vert\leq B \mbox{ for all }n\in\mbox{{\bf Z}}_{0\leq j\leq N_f(1)}\cup\mbox{{\bf Z}}_{\geq N_f(1)},$$

i.e., $\vert f(n)\vert\leq B$ for all $n\in\mbox{{\bf N}}$. $\mid\!\mid\!\mid$

7.60   Exercise. A Complete the proof of theorem 7.59; i.e., show that if $\{\alpha_n\}$ is a convergent complex sequence, then $\{\alpha_n\}$ is bounded.

7.61   Example. It follows from the fact that convergent sequences are bounded, that $\{n\}$ is not a convergent sequence.

7.62   Exercise. Give an example of a bounded sequence that is not convergent.