Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 9/refcontrol
- Series
We have seen in the last lecture that one can consider a number in the decimal numeral system, meaning an (infinite) sequence of digits between and , as an increasing sequence of rational numbers. For this, the -th digit after the separator, namely , means and has to be added to the approximation given by the digits before. The sequence of digits describes with the inverse powers of 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.
Definition
Let be a sequenceMDLD/sequence (R) of real numbers.MDLD/real numbers The series is the sequence of the partial sums
If the sequence converges,MDLD/converges (R) then we say that the series converges. In this case, we write also
for its limit,MDLD/limit (real sequence)
and this limit is called the sum of the series.All concepts for sequences carry over to series if we consider a series as the sequence of its partial sums . Like for sequences, it might happen that the sequence does not start with but later.
== Example Example 9.2
change==
We want to compute the series
For this, we give a formula for the -th partial sum. We have
This sequence converges to , so that the series converges and its sum equals .
LemmaCreate referencenumber
Let
denote convergent seriesMDLD/convergent series (R) of real numbersMDLD/real numbers with sums and
respectively. Then the following statements hold.- The series given by is also convergent and its sum is .
- For also the series given by is convergent and its sum is .
Proof
LemmaLemma 9.4 change
Let
be a seriesMDLD/series (R) of real numbers.MDLD/real numbers Then the series is convergentMDLD/convergent (series R) if and only if the following Cauchy-criterion holds: For every there exists some such that for all
the estimate
holds.
Proof
LemmaCreate referencenumber
Let
denote a convergentMDLD/convergent (series R) seriesMDLD/series (R) of real numbers.MDLD/real numbers Then
Proof
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 harmonic series shows.
== Example Example 9.6
change==
The harmonic series is the series
This series diverges: For the numbers , we have
Therefore,
Hence, the sequence of the partial sums is unbounded,MDLD/unbounded (sequence R) and so, due to Lemma 9.10 , not convergent.MDLD/convergent (sequence R)
The following statement is called Leibniz criterion for alternating series.
TheoremCreate referencenumber
Let be an decreasing null sequenceMDLD/null sequence of nonnegative real numbers.MDLD/real numbers Then the seriesMDLD/series (R) converges.MDLD/converges (series R)
Proof
- Absolutely convergent series
Definition
of real numbersMDLD/real numbers is called absolutely convergent, if the series
LemmaLemma 9.9 change
Proof
Let be given. We use the Cauchy-criterion. Since the series converges absolutely,MDLD/converges absolutely (R) there exists some such that for all the estimate
holds. Therefore,
which means the convergence.MDLD/convergence (series R)
Example Create referencenumber
A convergent series does not in general converge absolutely,MDLD/converge absolutely (R) the converse of Lemma 9.9 does not hold. Due to the Leibniz criterion, the alternating harmonic series
converges, and its sum is , 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 direct comparison test.
LemmaCreate referencenumber
Let be a convergent seriesMDLD/convergent series (R) of real numbersMDLD/real numbers and a sequenceMDLD/sequence of real numbersMDLD/real numbers fulfilling for all . Then the series
Proof
This follows directly from the Cauchy-criterion.
Example Create referencenumber
We want to determine whether the series
converges. We use the direct comparison test and Example 9.2 , where we have shown the convergence of . For we have
Hence, converges and therefore also . This does not say much about the exact value of the sum. With much more advanced methods, one can show that this sum equals .
- Geometric series and ratio test
The series is called geometric series for , so this is the sum
The convergence depends heavily on the modulus of .
TheoremTheorem 9.13 change
For all real numbersMDLD/real numbers with , the geometric seriesMDLD/geometric series (R) converges absolutely,MDLD/absolutely (convergence R) and the sum equals
Proof
For every and every we have the relation
and hence for the partial sumsMDLD/partial sums (R) the relation (for )
holds. For and this convergesMDLD/converges (R) to because of Lemma 8.1 and exercise *****.
The following statement is called ratio test.
TheoremCreate referencenumber
Let
be a seriesMDLD/series (R) of real numbers.MDLD/real numbers Suppose there exists a real numberMDLD/real number with , and a with
for all (in particular for ). Then the series converges absolutely.MDLD/converges absolutely (R)
Proof
The convergence does not change (though the sum) when we change finitely many members of the series. Therefore, we can assume . Moreover, we can assume that all are positiveMDLD/positive (R) real numbers.MDLD/real numbers Then
Hence, the convergence follows from the comparison test and the convergence of the geometric series.MDLD/geometric series (R)
Example Create referencenumber
The Koch snowflakes are given by the sequence of plane geometric shapes , which are defined recursively in the following way: The starting object is an equilateral triangle. The object is obtained from by replacing in each edge of the third in the middle by the corresponding equilateral triangle showing outside.
Let denote the area and the length of the boundary of the -th Koch snowflake. We want to show that the sequence convergesMDLD/converges (sequence R) and that the sequence diverges to .
The number of edges of is , since in each division step, one edge is replaced by four edges. Their length is of the length of a previous edge. Let denote the base length of the starting equilateral triangle. Then consists of edges of length and the length of all edges of together is
Because of , this diverges to .
When we turn from to , 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 is . So in the step from to there are triangles added with area . The total area of is therefore
If we forget the and the factor , which does not change the convergence property, we get in the bracket a partial sum of the geometric series for , and this converges.
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