Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 9/latex
\setcounter{section}{9}
\zwischenueberschrift{Series}
We have seen in the last lecture that one can consider a number in the decimal numeral system, meaning an \zusatzklammer {infinite} {} {} sequence of digits between \mathkor {} {0} {and} {9} {,} as an increasing sequence of rational numbers. For this, the $n$-th digit after the separator, namely \mathl{z_{-n}}{,} means \mathl{z_{-n} \cdot 10^{-n}}{} and has to be added to the approximation given by the digits before. The sequence of digits describes with the inverse powers of $10$ the difference between the approximating sequence, and the members in the approximating sequence are gained by summing up these differences. This viewpoint leads to the concept of a series.
\inputdefinition
{ }
{
Let \mathl{{ \left( a_k \right) }_{k \in \N }}{} be a
sequence
of
real numbers.
The \definitionswort {series}{} \mathl{\sum_{ k = 0}^\infty a_{ k }}{} is the sequence \mathl{{ \left( s_n \right) }_{n \in \N }}{} of the \definitionswort {partial sums}{}
\mavergleichskettedisp
{\vergleichskette
{ s_n
}
{ \defeq} { \sum_{ k = 0}^n a_{ k }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
If the sequence \mathl{{ \left( s_n \right) }_{n \in \N }}{}
converges,
then we say that the \definitionswort {series converges}{.} In this case, we write also
\mathdisp {\sum_{ k = 0}^\infty a_{ k }} { }
for its
limit,
}
All concepts for sequences carry over to series if we consider a series \mathl{\sum_{ k = 0}^\infty a_{ k }}{} as the sequence of its partial sums
\mavergleichskette
{\vergleichskette
{ s_n
}
{ = }{ \sum_{ k = 0}^n a_{ k }
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
Like for sequences, it might happen that the sequence does not start with
\mavergleichskette
{\vergleichskette
{k
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
but later.
\inputbeispiel{}
{
We want to compute the series
\mathdisp {\sum_{k=1}^\infty { \frac{ 1 }{ k(k+1) } }} { . }
For this, we give a formula for the $n$-th partial sum. We have
\mavergleichskettedisp
{\vergleichskette
{ s_n
}
{ =} { \sum_{k = 1}^n { \frac{ 1 }{ k(k+1) } }
}
{ =} { \sum_{k = 1}^n { \left( { \frac{ 1 }{ k } } - { \frac{ 1 }{ k+1 } } \right) }
}
{ =} { 1- { \frac{ 1 }{ n+1 } }
}
{ =} { { \frac{ n }{ n+1 } }
}
}
{}{}{.}
This sequence converges to $1$, so that the series converges and its sum equals $1$.
}
\inputfaktbeweis
{Real numbers/Series/Rules/Fact}
{Lemma}
{}
{
\faktsituation {Let
\mathdisp {\sum_{ k = 0}^\infty a_{ k } \text{ and } \sum_{ k = 0}^\infty b_{ k }} { }
denote
convergent series
of
real numbers
with sums
\mathkor {} {s} {and} {t} {}
respectively.}
\faktuebergang {Then the following statements hold.}
\faktfolgerung {\aufzaehlungzwei {The series \mathl{\sum_{ k = 0}^\infty c_{ k }}{} given by
\mavergleichskette
{\vergleichskette
{ c_k
}
{ \defeq }{ a_k+b_k
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
is also convergent and its sum is \mathl{s+t}{.}
} {For
\mavergleichskette
{\vergleichskette
{ r
}
{ \in }{ \R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
also the series \mathl{\sum_{ k = 0}^\infty d_{ k }}{} given by
\mavergleichskette
{\vergleichskette
{ d_k
}
{ \defeq }{ r a_k
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
is convergent and its sum is $r s$.
}}
\faktzusatz {}
{See Exercise 9.8 . }
\inputfaktbeweis
{Real_numbers/Series/Cauchy-criterion/Fact}
{Lemma}
{}
{
\faktsituation {Let
\mathdisp {\sum_{ k = 0}^\infty a_{ k }} { }
be a
series
of
real numbers.}
\faktfolgerung {Then the series is
convergent
if and only if the following \stichwort {Cauchy-criterion} {} holds: For every
\mavergleichskette
{\vergleichskette
{\epsilon
}
{ > }{0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
there exists some $n_0$ such that for all
\mavergleichskettedisp
{\vergleichskette
{n
}
{ \geq} {m
}
{ \geq} {n_0
}
{ } {
}
{ } {
}
}
{}{}{}
the estimate
\mavergleichskettedisp
{\vergleichskette
{ \betrag { \sum_{k = m}^n a_k }
}
{ \leq} { \epsilon
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
holds.}
\faktzusatz {}
{See Exercise 9.9 . }
\inputfaktbeweis
{Real numbers/Series converges/Summands/Null sequence/Fact}
{Lemma}
{}
{
\faktsituation {Let
\mathdisp {\sum_{ k = 0}^\infty a_{ k }} { }
denote a
convergent
series
of
real numbers.}
\faktfolgerung {Then
\mavergleichskettedisp
{\vergleichskette
{ \lim_{ k \rightarrow \infty} a_{ k }
}
{ =} { 0
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}}
\faktzusatz {}
}
{
This follows directly from Lemma 9.4 .
It is therefore a necessary condition for the convergence of a series that its members form a null sequence. This condition is not sufficient, as the \stichwort {harmonic series} {} shows.
\inputbeispiel{}
{
The \definitionswort {harmonic series}{} is the series
\mathdisp {\sum^\infty_{k = 1} { \frac{ 1 }{ k } }} { . }
\mathdisp {1 + { \frac{ 1 }{ 2 } } + { \frac{ 1 }{ 3 } } + { \frac{ 1 }{ 4 } } + { \frac{ 1 }{ 5 } } + { \frac{ 1 }{ 6 } } + { \frac{ 1 }{ 7 } } + { \frac{ 1 }{ 8 } } + \ldots} { . }
This series diverges: For the $2^{n}$ numbers
\mavergleichskette
{\vergleichskette
{ k
}
{ = }{ 2^n +1 , \ldots , 2^{n+1}
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
we have
\mavergleichskettedisp
{\vergleichskette
{ \sum_{k = 2^n+1}^{ 2^{n+1} } \frac{1}{k}
}
{ \geq} { \sum_{k = 2^n+1}^{ 2^{n+1} } \frac{1}{2^{n+1} }
}
{ =} { 2^n \frac{1}{2^{n+1} }
}
{ =} { \frac{1}{2}
}
{ } {
}
}
{}{}{.}
Therefore,
\mavergleichskettedisp
{\vergleichskette
{ \sum_{k = 1}^{ 2^{n+1} } \frac{1}{k}
}
{ =} {1+ \sum_{i = 0}^n \left( \sum_{k = 2^{i} +1 }^{ 2^{i+1} } \frac{1}{k} \right)}
{ } {
}
{ \geq} {1 + (n+1) \frac{1}{2}
}
{ } {
}
}
{}{}{.}
Hence, the sequence of the partial sums is
unbounded,
and so, due to
Lemma 9.10
,
not
convergent.
}
The following statement is called \stichwort {Leibniz criterion for alternating series} {.}
\inputfaktbeweisnichtvorgefuehrt
{Series/Real numbers/Leibniz criterion/Fact}
{Theorem}
{}
{
\faktsituation {Let \mathl{{ \left( x_k \right) }_{k \in \N }}{} be an decreasing
null sequence
of nonnegative
real numbers.}
\faktfolgerung {Then the
series
\mathl{\sum_{ k= 0}^\infty (-1)^k x_k}{}
converges.}
\faktzusatz {}
\zwischenueberschrift{Absolutely convergent series}
\inputdefinition
{ }
{
A
series
\mathdisp {\sum_{ k = 0}^\infty a_{ k }} { }
of
real numbers
is called \definitionswort {absolutely convergent}{,} if the series
\mathdisp {\sum_{ k = 0}^\infty \betrag { a_k }} { }
}
\inputfaktbeweis
{Real series/Absolute convergence and convergence/Fact}
{Lemma}
{}
{
\faktsituation {}
\faktvoraussetzung {An
absolutely convergent
series
of
real numbers}
\faktfolgerung {converges.}
\faktzusatz {}
}
{
Let
\mavergleichskette
{\vergleichskette
{ \epsilon
}
{ > }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
be given. We use the
Cauchy-criterion.
Since the series
converges absolutely,
there exists some $n_0$ such that for all
\mavergleichskette
{\vergleichskette
{n
}
{ \geq }{ m
}
{ \geq }{ n_0
}
{ }{
}
{ }{
}
}
{}{}{}
the estimate
\mavergleichskettedisp
{\vergleichskette
{ \betrag { \sum_{k = m}^n \betrag { a_k } \, }
}
{ =} { \sum_{k = m}^n \betrag { a_k }
}
{ \leq} { \epsilon
}
{ } {
}
{ } {
}
}
{}{}{}
holds. Therefore,
\mavergleichskettedisp
{\vergleichskette
{ \betrag { \sum_{k = m}^n a_k }
}
{ \leq} { \betrag { \sum_{k = m}^n \betrag { a_k } \, }
}
{ \leq} { \epsilon
}
{ } {
}
{ } {
}
}
{}{}{,}
which means the
convergence.
\inputbeispiel{}
{
A convergent series does not in general
converge absolutely,
the converse of
Lemma 9.9
does not hold. Due to the
Leibniz criterion,
the \stichwort {alternating harmonic series} {}
\mavergleichskettedisp
{\vergleichskette
{ \sum_{n = 1}^\infty \frac{ (-1)^{n+1} }{n}
}
{ =} { 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
converges, and its sum is $\ln 2$, a result we can not prove here. However, the corresponding absolute series is just the harmonic series, which diverges due to
Example 9.6
.
}
The following statement is called the \stichwort {direct comparison test} {.}
\inputfaktbeweis
{Real series/Direct comparison test/Fact}
{Lemma}
{}
{
\faktsituation {}
\faktvoraussetzung {Let \mathl{\sum_{ k = 0}^\infty b_{ k }}{} be a
convergent series
of
real numbers
and \mathl{{ \left( a_k \right) }_{k \in \N }}{} a
sequence
of
real numbers
fulfilling
\mavergleichskette
{\vergleichskette
{ \betrag { a_k }
}
{ \leq }{ b_k
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
for all $k$.}
\faktfolgerung {Then the series
\mathdisp {\sum_{ k = 0}^\infty a_{ k }} { }
is
absolutely convergent.}
\faktzusatz {}
}
{
This follows directly from the Cauchy-criterion.
\inputbeispiel{}
{
We want to determine whether the series
\mavergleichskettedisp
{\vergleichskette
{ \sum_{k = 1}^\infty { \frac{ 1 }{ k^2 } }
}
{ =} { 1 + { \frac{ 1 }{ 4 } } + { \frac{ 1 }{ 9 } } + { \frac{ 1 }{ 16 } } + { \frac{ 1 }{ 25 } } + \ldots
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
converges. We use
the direct comparison test
and
Example 9.2
,
where we have shown the convergence of \mathl{\sum_{k = 1}^n { \frac{ 1 }{ k(k+1) } }}{.} For
\mavergleichskette
{\vergleichskette
{k
}
{ \geq }{2
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
we have
\mavergleichskettedisp
{\vergleichskette
{ { \frac{ 1 }{ k^2 } }
}
{ \leq} { { \frac{ 1 }{ k(k-1) } }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
Hence, \mathl{\sum_{k=2}^\infty { \frac{ 1 }{ k^2 } }}{} converges and therefore also \mathl{\sum_{k=1}^\infty { \frac{ 1 }{ k^2 } }}{.} This does not say much about the exact value of the sum. With much more advanced methods, one can show that this sum equals \mathl{{ \frac{ \pi^2 }{ 6 } }}{.}
}
\zwischenueberschrift{Geometric series and ratio test}
\mavergleichskette
{\vergleichskette
{ x }
{ = }{ { \frac{ 1 }{ 4 } } }
{ }{ }
{ }{ }
{ }{ }
} {}{}{.} Suppose that the length of the square is $2$. Then we can put the geometric series threetimes inside this square. The area of the three series is ${ \frac{ 4 }{ 3 } }$.
The series \mathl{\sum_{ k = 0}^\infty x^k}{} is called \stichwort {geometric series} {} for
\mavergleichskette
{\vergleichskette
{x
}
{ \in }{ \R
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
so this is the sum
\mathdisp {1+x+x^2+x^3+ \ldots} { . }
The convergence depends heavily on the modulus of $x$.
\inputfaktbeweis
{Geometric series/Real/Convergence/Absolutely/Fact}
{Theorem}
{}
{
For all
real numbers
$x$ with
\mavergleichskette
{\vergleichskette
{ \betrag { x }
}
{ < }{1
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
the
geometric series
\mathl{\sum^\infty_{k = 0} x^k}{} converges
absolutely,
and the sum equals
\mavergleichskettedisp
{\vergleichskette
{ \sum^\infty_{k = 0} x^k
}
{ =} { { \frac{ 1 }{ 1-x } }
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{.}
}
{
For every $x$ and every
\mavergleichskette
{\vergleichskette
{n
}
{ \in }{ \N
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
we have the relation
\mavergleichskettedisp
{\vergleichskette
{ (x-1) { \left(\sum_{ k = 0}^n x^k\right) }
}
{ =} { x^{n+1} -1
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
and hence for the
partial sums
the relation (for
\mavergleichskette
{\vergleichskette
{ x
}
{ \neq }{ 1
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
)
\mavergleichskettedisp
{\vergleichskette
{ s_n
}
{ =} { \sum_{ k = 0}^n x^k
}
{ =} { \frac{ x^{n+1} -1}{ x-1 }
}
{ } {
}
{ } {}
}
{}{}{}
holds. For \mathl{n \rightarrow \infty}{} and
\mavergleichskette
{\vergleichskette
{ \betrag { x }
}
{ < }{ 1
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
this
converges
to
\mavergleichskette
{\vergleichskette
{ \frac{-1}{x-1}
}
{ = }{ \frac{1}{1-x}
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
because of
Lemma 8.1
and
exercise *****.
The following statement is called \stichwort {ratio test} {.}
\inputfaktbeweis
{Real series/Ratio test/Fact}
{Theorem}
{}
{
\faktsituation {Let
\mathdisp {\sum_{ k = 0}^\infty a_{ k }} { }
be a
series
of
real numbers.}
\faktvoraussetzung {Suppose there exists a
real number
$q$ with
\mavergleichskette
{\vergleichskette
{ 0
}
{ \leq }{ q
}
{ < }{ 1
}
{ }{
}
{ }{
}
}
{}{}{,}
and a $k_0$ with
\mavergleichskettedisp
{\vergleichskette
{ \betrag { { \frac{ a_{k+1} }{ a_k } } }
}
{ \leq} { q
}
{ } {
}
{ } {
}
{ } {
}
}
{}{}{}
for all
\mavergleichskette
{\vergleichskette
{ k
}
{ \geq }{ k_0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
\zusatzklammer {in particular
\mavergleichskette
{\vergleichskette
{ a_k
}
{ \neq }{0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}
for
\mavergleichskette
{\vergleichskette
{ k
}
{ \geq }{ k_0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{}} {} {.}}
\faktfolgerung {Then the series \mathl{\sum_{ k = 0}^\infty a_{ k }}{}
converges absolutely.}
\faktzusatz {}
}
{
The convergence does not change
\zusatzklammer {though the sum} {} {}
when we change finitely many members of the series. Therefore, we can assume
\mavergleichskette
{\vergleichskette
{ k_0
}
{ = }{ 0
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
Moreover, we can assume that all $a_k$ are
positive
real numbers.
Then
\mavergleichskettedisp
{\vergleichskette
{ a_k
}
{ =} { \frac{a_k}{a_{k-1} } \cdot \frac{a_{k-1} }{a_{k-2} } \cdots \frac{a_1}{a_{0} } \cdot a_0
}
{ \leq} { a_0 \cdot q^k
}
{ } {
}
{ } {
}
}
{}{}{.}
Hence, the convergence follows from the
comparison test
and the
convergence
of the
geometric series.
\inputbeispiel{}
{
The \stichwort {Koch snowflakes} {} are given by the sequence of plane geometric shapes $K_n$, which are defined recursively in the following way: The starting object $K_0$ is an equilateral triangle. The object $K_{n+1}$ is obtained from $K_n$ by replacing in each edge of $K_n$ the third in the middle by the corresponding equilateral triangle showing outside.
Let $A_n$ denote the area and $L_n$ the length of the boundary of the $n$-th Koch snowflake. We want to show that the sequence $A_n$ converges and that the sequence $L_n$ diverges to $\infty$.
The number of edges of $K_n$ is \mathl{3 \cdot 4^n}{,} since in each division step, one edge is replaced by four edges. Their length is \mathl{1/3}{} of the length of a previous edge. Let $r$ denote the base length of the starting equilateral triangle. Then $K_n$ consists of \mathl{3 \cdot 4^n}{} edges of length \mathl{r { \left( { \frac{ 1 }{ 3 } } \right) }^n}{} and the length of all edges of $K_n$ together is
\mavergleichskettedisp
{\vergleichskette
{ L_n
}
{ =} { 3 \cdot 4^n r { \left( { \frac{ 1 }{ 3 } } \right) }^n
}
{ =} { 3 r { \left( { \frac{ 4 }{ 3 } } \right) }^n
}
{ } {
}
{ } {
}
}
{}{}{.}
Because of
\mavergleichskette
{\vergleichskette
{ { \frac{ 4 }{ 3 } }
}
{ > }{ 1
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{,}
this diverges to $\infty$.
When we turn from $K_{n}$ to \mathl{K_{n+1}}{,} there will be for every edge a new triangle whose side length is a third of the edge length. The area of an equilateral triangle with side length $s$ is \mathl{{ \frac{ \sqrt{3} }{ 4 } } s^2}{.} So in the step from \mathl{K_{n}}{} to \mathl{K_{n+1}}{} there are \mathl{3 \cdot 4^{n}}{} triangles added with area
\mavergleichskette
{\vergleichskette
{ { \frac{ \sqrt{3} }{ 4 } } { \left( { \frac{ 1 }{ 3 } } \right) }^{2(n+1)} r^2
}
{ = }{ { \frac{ \sqrt{3} }{ 4 } } r^2 { \left( { \frac{ 1 }{ 9 } } \right) }^{n+1}
}
{ }{
}
{ }{
}
{ }{
}
}
{}{}{.}
The total area of $K_n$ is therefore
\mavergleichskettealign
{\vergleichskettealign
{
}
{ \,} { { \frac{ \sqrt{3} }{ 4 } } r^2 { \left( 1 + 3 { \frac{ 1 }{ 9 } } + 12 { \left( { \frac{ 1 }{ 9 } } \right) } ^2 + 48 { \left( { \frac{ 1 }{ 9 } } \right) }^3 + \cdots + 3\cdot 4^{n-1} { \left( { \frac{ 1 }{ 9 } } \right) }^{n} \right) }
}
{ =} { { \frac{ \sqrt{3} }{ 4 } } r^2 { \left( 1 + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^1 + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^2 + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^3 + \cdots + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^{n} \right) }
}
{ } {
}
{ } {
}
}
{}
{}{.}
If we forget the $1$ and the factor \mathl{{ \frac{ 3 }{ 4 } }}{,} which does not change the convergence property, we get in the bracket a partial sum of the geometric series for ${ \frac{ 4 }{ 9 } }$, and this converges.
}