Next: 7.6 Geometric Series
Up: 7. Complex Sequences
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Proof: By definition 7.10, ``
" means
for every
![$r\in\mbox{${\mbox{{\bf R}}}^{+}$}$](img68.gif)
there
is some
![$N\in\mbox{{\bf N}}$](img31.gif)
such that
for every
![$n\in\mbox{{\bf Z}}_{\geq N}$](img71.gif)
,
![$\left(\vert f(n)-L\vert<r\right)$](img198.gif)
.
By definition 7.11, ``
is a null sequence" means
Both definitions say the same thing. If we write out the definition for ``
is a null sequence" we get (7.30) with ``
" replaced by
``
." Since
conditions b) and c) are equivalent.
Proof:
.
Proof: Suppose
and
.
By the decomposition theorem, we can write
where
and
are null sequences. Then
By the sum theorem for null sequences,
is a null sequence, so
is a null sequence, and hence
.
7.33
Exercise.
Prove the last statement in theorem
7.32; i.e., show that if
![$f\to L$](img29.gif)
then
![$\alpha f\to\alpha L$](img218.gif)
for all
![$\alpha\in\mbox{{\bf C}}$](img10.gif)
.
7.34
Theorem (Product theorem for convergent sequences.)
Let
be convergent complex sequences. Suppose
and
. Then
is
convergent and
.
Proof: Suppose
and
. Write
,
where
are null sequences. Then
Now
,
and
are null sequences by the product theorem and sum theorem for
null sequences, and
, so by several applications of the sum
theorem for convergent sequences,
7.35
Theorem (Uniqueness theorem for convergent sequences.)
Let
be a complex sequence, and let
. If
and
, then
.
Proof: Suppose
and
. Then
and
are null
sequences, so
is a null
sequence. Hence, by theorem 7.15,
; i.e.,
.
7.36
Definition (Limit of a sequence.)
Let
![$f$](img13.gif)
be a convergent sequence. Then the unique complex number
![$L$](img28.gif)
such that
![$f\to L$](img29.gif)
is denoted by
![$\lim f$](img235.gif)
or
![$\lim\{f(n)\}$](img236.gif)
.
7.37
Remark.
It follows from the sum and product theorems that if
![$f$](img13.gif)
and
![$g$](img111.gif)
are convergent
sequences, then
and
and
7.38
Warning.
We have only defined
![$\lim f$](img235.gif)
when
![$f$](img13.gif)
is a convergent sequence. Hence
![$\lim\{i^n\}$](img240.gif)
is ungrammatical and should not be written down. (We showed in theorem
7.7 that
![$\{i^n\}$](img241.gif)
diverges.) However, it is a standard
usage to say ``
![$\lim f$](img235.gif)
does not exist'' or ``
![$\lim\{f(n)\}$](img236.gif)
does
not exist'' to mean that the sequence
![$f$](img13.gif)
has no limit. Hence we may say ``
![$\lim\{i^n\}$](img240.gif)
does not exist''.
Proof: If
and
are convergent, then it follows from the sum
theorem for convergent sequences that
is convergent and (7.40) is valid.
Suppose that
. Then
is a null sequence, so
is a null sequence (by Theorem 7.26). For all
,
so
is a null sequence and it
follows that
converges to
. A similar argument shows that
.
7.41
Definition (Bounded sequence.)
A sequence
![$f$](img13.gif)
in
![$\mbox{{\bf C}}$](img2.gif)
is
bounded, if there is a disc
![$\overline{D}(0,B)$](img254.gif)
such that
![$f(n)\in\overline D(0,B)$](img255.gif)
for all
![$n\in\mbox{{\bf N}}$](img9.gif)
; i.e.,
![$f$](img13.gif)
is bounded if there is a number
![$B\in[0,\infty)$](img256.gif)
such that
![\begin{displaymath}
\vert f(n)\vert\leq B \mbox{ for all } n\in\mbox{{\bf N}}.
\end{displaymath}](img257.gif) |
(7.42) |
Any number
![$B$](img258.gif)
satisfying condition (
7.42) is called a
bound for ![$f$](img13.gif)
.
7.43
Examples.
![$\displaystyle {\left\{ {{i^nn}\over {n+1}}\right\}}$](img259.gif)
is bounded since
![$\displaystyle {\left\vert{{i^nn}\over {n+1}}\right\vert={n\over {n+1}}\leq 1}$](img260.gif)
for all
![$n\in\mbox{{\bf N}}$](img9.gif)
. The
sequence
![$\{n\}$](img261.gif)
is not bounded since the statement
![$\vert n\vert\leq B$](img262.gif)
for all
![$n\in\mbox{{\bf N}}$](img9.gif)
contradicts the Archimedean property of
![$\mbox{{\bf R}}$](img4.gif)
. Every constant sequence
![$\{\tilde
L\}$](img263.gif)
is bounded. In fact,
![$\vert L\vert$](img264.gif)
is a bound for
![$\tilde L$](img265.gif)
.
7.44
Exercise (Null-times-bounded theorem.)
A
Show that if
![$f$](img13.gif)
is a null sequence in
![$\mbox{{\bf C}}$](img2.gif)
, and
![$g$](img111.gif)
is a bounded sequence in
![$\mbox{{\bf C}}$](img2.gif)
then
![$fg$](img195.gif)
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
, then
Proof: By the triangle inequality.
Hence,
7.46
Lemma.
Let
be a convergent sequence
that is not a null sequence; i.e.,
where
. Suppose
for all
. Then
is a bounded sequence.
Proof: Since
, we know that
is a null sequence. Let
be a
precision function for
. Then for all
,
i.e., if
, then
Let
Then
for
and
for
, so
for all
, and hence
is bounded.
7.47
Theorem (Reciprocal theorem for convergent sequences.)
Let
be a complex sequence. Suppose that
where
, and that
for all
. Then
is convergent, and
.
Proof: By the preceding lemma,
is a bounded sequence, and since
, we know that
is a null sequence. Hence
is a null sequence, and it follows that
. Then we have
i.e.,
.
7.49
Exercise.
A
- a)
- Let
be complex sequences.
Show that if
converges
and
diverges, then
diverges.
- b)
- Show that if
converges and
diverges, then
does not necessarily
diverge.
7.50
Exercise.
A
Let
![$f$](img13.gif)
be a divergent complex sequence.
Show that if
![$c\in\mbox{{\bf C}}\backslash\{0\}$](img292.gif)
, then
![$cf$](img293.gif)
is divergent.
7.51
Example.
Let
![$f\colon\mbox{{\bf Z}}_{\geq 1}\to\mbox{{\bf C}}$](img294.gif)
be defined by
![\begin{displaymath}
{f(n)={{n^2+in+1}\over {3n^2+2in-1}}}.
\end{displaymath}](img295.gif) |
(7.52) |
Then
![\begin{displaymath}
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}}}.
\end{displaymath}](img296.gif) |
(7.53) |
Hence
![$f$](img13.gif)
can be written as a quotient of two sequences:
and
where
![$g(n)\neq 0$](img283.gif)
for all
![$n\in\mbox{{\bf Z}}_{\geq 1}$](img117.gif)
.
Since
and
it
follows from numerous applications of product and sum rules that
![$h\to 1$](img301.gif)
and
![$g\to
3\neq 0$](img302.gif)
and hence
![$\displaystyle {f={h\over g}\to{1\over 3}}$](img303.gif)
. Once I have expressed
![$f(n)$](img304.gif)
in
the final form in (
7.53), I can see what the final result is, and I will usually
just write
7.54
Example.
Let
![$g\colon\mbox{{\bf N}}\to\mbox{{\bf C}}$](img306.gif)
be the sequence
![\begin{displaymath}
g=\left\{{{2^n+4^n}\over {4^n+6^n}}\right\}.
\end{displaymath}](img307.gif) |
(7.55) |
Then for all
![$n\in\mbox{{\bf N}}$](img9.gif)
,
Since
![$\displaystyle {\left\vert{2\over 3}\right\vert<.7}$](img309.gif)
, I know
![$\displaystyle {\left\{\left({2\over
3}\right)^n\right\}\to 0}$](img310.gif)
and
![$\displaystyle {\left\{\left({1\over 2}\right)^n\right\}\to 0}$](img311.gif)
so
In the last two examples, I was motivated by the following considerations. I think:
In the numerator and denominator for (7.52), for large
the ``
"
term overwhelms the other terms - so that's the term I factored out. In the
numerator of (7.55), the overwhelming term is
, and in the denominator,
the overwhelming term is
so those are the terms I factored out.
7.56
Exercise.
A
Let
![$\{f(n)\}$](img317.gif)
be a sequence of non-negative numbers
and suppose
![$\{f(n)\}\to L$](img318.gif)
where
![$L>0$](img319.gif)
. Prove that
![$\displaystyle {\{\sqrt{f(n)}\}\to\sqrt L}$](img320.gif)
. (NOTE:
The case
![$L=0$](img321.gif)
follows from the root theorem for null sequences.
7.58
Exercise.
A
Show that the sum of two
bounded sequences is a bounded
sequence.
7.59
Theorem (Convergent sequences are bounded.)
Let
be a convergent complex sequence. Then
is bounded.
Proof: I will show that null sequences are bounded and leave the general case to
you. Let
be a null sequence and let
be a precision function for
.
Let
I claim that
is a bound for
. If
, then
If
, then
, so
. Hence
i.e.,
for all
.
7.60
Exercise.
A
Complete the proof of theorem
7.59; i.e., show that if
![$\{\alpha_n\}$](img327.gif)
is a convergent complex sequence, then
![$\{\alpha_n\}$](img327.gif)
is bounded.
7.61
Example.
It follows from the fact that convergent sequences are bounded, that
![$\{n\}$](img261.gif)
is not a convergent sequence.
7.62
Exercise.
Give an example of a bounded sequence that is not convergent.
Next: 7.6 Geometric Series
Up: 7. Complex Sequences
Previous: 7.4 Sums and Products
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