Lectures 1 & 2 : Sequences and Series...Lectures 1 & 2 : Sequences and Series Prof. Kurt Helmes...
Transcript of Lectures 1 & 2 : Sequences and Series...Lectures 1 & 2 : Sequences and Series Prof. Kurt Helmes...
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Lectures 1 & 2 :
Sequences and Series
Prof. Kurt Helmes
Institute of Operations Research
Humboldt-University of Berlin
6th Summer-School Havanna - ISSEM 2008
Sep. 22th - Oct. 2th
Prof. Kurt Helmes Sequences and Series
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Table of content
Part 1: Sequences
Part 2: Series
Prof. Kurt Helmes Sequences and Series
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Table of content
Part 1: Sequences
Part 2: Series
Prof. Kurt Helmes Sequences and Series
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1 Sequences
Prof. Kurt Helmes Sequences and Series
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Part 1.1
Definition of Sequences
Prof. Kurt Helmes Sequences and Series
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Definition: Sequences
A sequence is a function, which is only “fed”with natural numbers:
n 7→ a(n) =: an n ∈ N0 = {0, 1, 2, 3, . . . }
(a possibility: n is an element of an infinite subset of N)
Prof. Kurt Helmes Sequences and Series
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Definition: Sequences
A sequence is a function, which is only “fed”with natural numbers:
n 7→ a(n) =: an n ∈ N0 = {0, 1, 2, 3, . . . }
(a possibility: n is an element of an infinite subset of N)
Prof. Kurt Helmes Sequences and Series
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Remark:
In many economic andbusiness type applications the
variable n is understood to referto a time point/index.
Prof. Kurt Helmes Sequences and Series
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Ways to describe sequences:
• by a table of values:table of values(a1, a2, a3, ..., an)
• by a transformation ruleww� ww�explicit recursive
Prof. Kurt Helmes Sequences and Series
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Ways to describe sequences:
• by a table of values:table of values(a1, a2, a3, ..., an)
• by a transformation ruleww� ww�explicit recursive
Prof. Kurt Helmes Sequences and Series
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Examples of explicit definitions:
an = a(n) = n2, n = 0, 1, 2, 3, . . .
Prof. Kurt Helmes Sequences and Series
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Examples of explicit definitions:
an = a(n) = n2, n = 0, 1, 2, 3, . . .
Prof. Kurt Helmes Sequences and Series
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Examples of explicit definitions:
Growth of an initial capital K0,with compound interest; the annualinterest rate is p%.
Kn =(1 + p
100
)n · K0, n = 0, 1, 2, 3, . . .
Prof. Kurt Helmes Sequences and Series
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Examples of explicit definitions:
Growth of an initial capital K0,with compound interest; the annualinterest rate is p%.
Kn =(1 + p
100
)n · K0, n = 0, 1, 2, 3, . . .
Prof. Kurt Helmes Sequences and Series
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Examples of explicit definitions:
f .i . : Kn =(
1 +p
100
)n· K0, n = 0, 1, 2, 3, . . .
K0 = 100p = 7 %yKn = 100 · 1, 07n
n = 0, 1, 2, ..., 24
Prof. Kurt Helmes Sequences and Series
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Examples of explicit definitions:
f .i . : Kn =(
1 +p
100
)n· K0, n = 0, 1, 2, 3, . . .
K0 = 100p = 7 %yKn = 100 · 1, 07n
n = 0, 1, 2, ..., 24
Prof. Kurt Helmes Sequences and Series
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Examples of explicit definitions:
f .i . : Kn =(
1 +p
100
)n· K0, n = 0, 1, 2, 3, . . .
K0 = 100p = 7 %yKn = 100 · 1, 07n
n = 0, 1, 2, ..., 24
Prof. Kurt Helmes Sequences and Series
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Examples of a recursive definition:
an+1 = −0.9an + 0.6, a0 = 1, n = 0, 1, 2, 3, . . .
Prof. Kurt Helmes Sequences and Series
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Examples of a recursive definition:
an+1 = −0.9an + 0.6, a0 = 1, n = 0, 1, 2, 3, . . .
Prof. Kurt Helmes Sequences and Series
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Examples of a recursive definition:
(see Newton’s method, lectures 5&6):
The sequence of numbers (an)n≥0 whichis definded by the formula:
an+1 = an
2 + 1an
a0 6= 0 any fixed number
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Examples of tables of values:
The gross national product of Germany 1980 - 1990:
Year real GDP (in Trillion DM)
1980 2.026
1981 2.026
1982 2.004
1983 2.045
1984 2.108
1985 2.149
1986 2.199
1987 2.233
1988 2.314
1989 2.411
1990 2.544Source: Deutsche Bundesbank
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Examples of tables of values:
The gross national product of Germany 1980 - 1990:
Year real GDP (in Trillion DM)
1980 2.026
1981 2.026
1982 2.004
1983 2.045
1984 2.108
1985 2.149
1986 2.199
1987 2.233
1988 2.314
1989 2.411
1990 2.544Source: Deutsche Bundesbank
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Examples of tables of values:
The values of the German stock index for 1999;d1 = Jan.-DAX1999 is the first value of the finite sequence.
Januar Februar Marz April Mai Juni Juli August September Oktober November
5137 5313 4968 4936 5470 5132 5560 5170 5432 5343 5702
(di )i≥1
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Examples of tables of values:
The values of the German stock index for 1999;d1 = Jan.-DAX1999 is the first value of the finite sequence.
Januar Februar Marz April Mai Juni Juli August September Oktober November
5137 5313 4968 4936 5470 5132 5560 5170 5432 5343 5702
(di )i≥1
Prof. Kurt Helmes Sequences and Series
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Part 1.2
Arithmetic Sequences
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Definition: Arithmetic sequences
An arithmetic sequence is a linear (affine) functionwhose domain is N0, N resp.
an = a0 + nd n = 0, 1, 2, 3, . . . ,
where:
a0 : y−intercept
n : variable
d : slope
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Definition: Arithmetic sequences
An arithmetic sequence is a linear (affine) functionwhose domain is N0, N resp.
an = a0 + nd n = 0, 1, 2, 3, . . . ,
where:
a0 : y−intercept
n : variable
d : slope
Prof. Kurt Helmes Sequences and Series
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Example: Arithmetic sequence
an = 1 + 2nya0 = 1d = 2n = 0, 1, 2, . . .
Prof. Kurt Helmes Sequences and Series
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Representation: Arithmetic sequences
Explicit form: an = a0 + nd
Recursive form: an+1 − an = d⇐⇒
(a0 is given) an+1 = an + d
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Representation: Arithmetic sequences
Explicit form: an = a0 + nd
Recursive form: an+1 − an = d⇐⇒
(a0 is given) an+1 = an + d
Prof. Kurt Helmes Sequences and Series
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Part 1.3
Geometric Sequences
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Definition: Geometric sequences
Geometric sequences are exponential functionswhose domain is restricted to N0, N resp.
an = a0qn n = 0, 1, 2, 3, . . .
where:
a0 : y−intercept
n : variable
q : “base ′′/multiplier
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Definition: Geometric sequences
Geometric sequences are exponential functionswhose domain is restricted to N0, N resp.
an = a0qn n = 0, 1, 2, 3, . . .
where:
a0 : y−intercept
n : variable
q : “base ′′/multiplier
Prof. Kurt Helmes Sequences and Series
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Example: Geometric sequence
an = 2nya0 = 1q = 2n = 0, 1, 2, . . .
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Representation: Geometric Sequences
Explicit form: an = a0qn
Recursive form: an+1 = qan
an+1
an= q ⇐⇒
(a0 6= 0 is given) an+1 − qan = 0
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Representation: Geometric Sequences
Explicit form: an = a0qn
Recursive form: an+1 = qan
an+1
an= q ⇐⇒
(a0 6= 0 is given) an+1 − qan = 0
Prof. Kurt Helmes Sequences and Series
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Representation: Geometric Sequences
Explicit form: an = a0qn
Recursive form: an+1 = qan
an+1
an= q ⇐⇒
(a0 6= 0 is given) an+1 − qan = 0
Prof. Kurt Helmes Sequences and Series
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Part 1.4
Properties of Sequences
Prof. Kurt Helmes Sequences and Series
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Concept: Bounded Sequences
A sequence is called bounded if and only if (= iff):
|an| ≤ const., n = 1, 2, 3, . . .
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Example of a bounded sequence
an = (−1)n 1n , n = 1, 2, . . .
yis a
boundedsequence|(−1)n 1
n | ≤ 1
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Example of a bounded sequence
an = (−1)n 1n , n = 1, 2, . . .
yis a
boundedsequence|(−1)n 1
n | ≤ 1
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Example of a bounded sequence
an = (−1)n 1n , n = 1, 2, . . .
yis a
boundedsequence|(−1)n 1
n | ≤ 1
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Example of an unbounded sequence:
an = n2, n = 1, 2, . . .
yis an
unboundedsequence
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Example of an unbounded sequence:
an = n2, n = 1, 2, . . .
yis an
unboundedsequence
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Example of an unbounded sequence:
an = n2, n = 1, 2, . . .
yis an
unboundedsequence
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Monotone increasing / decreasing sequences:
monotone increasing:an ≤ an+1x ⇔
an+1 − an ≥ 0n = 1, 2, . . .
monotone decreasing:an ≥ an+1y ⇔
an+1 − an ≤ 0n = 1, 2, . . .
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Monotone increasing / decreasing sequences:
monotone increasing:an ≤ an+1x ⇔
an+1 − an ≥ 0n = 1, 2, . . .
monotone decreasing:an ≥ an+1y ⇔
an+1 − an ≤ 0n = 1, 2, . . .
Prof. Kurt Helmes Sequences and Series
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Monotone increasing / decreasing sequences:
monotone increasing:an ≤ an+1x ⇔
an+1 − an ≥ 0n = 1, 2, . . .
monotone decreasing:an ≥ an+1y ⇔
an+1 − an ≤ 0n = 1, 2, . . .
Prof. Kurt Helmes Sequences and Series
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Example: Monotone increasing
an = n2, n = 1, 2, . . .
yis a
monotoneincreasingsequence
(n + 1)2 =n2 + 2n+1≥ n2
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Example: Monotone increasing
an = n2, n = 1, 2, . . .yis a
monotoneincreasingsequence
(n + 1)2 =n2 + 2n+1≥ n2
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Example: Monotone increasing
an = n2, n = 1, 2, . . .yis a
monotoneincreasingsequence
(n + 1)2 =n2 + 2n+1≥ n2
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Example: Monotone increasing
an = 1n , n = 1, 2, . . .
yis a
monotonedecreasingsequence
1(n+1) =
nn+1 ·
1n + 2n+1
≤ 1n
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Example: Monotone increasing
an = 1n , n = 1, 2, . . .yis a
monotonedecreasingsequence
1(n+1) =
nn+1 ·
1n + 2n+1
≤ 1n
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Example: Monotone increasing
an = 1n , n = 1, 2, . . .yis a
monotonedecreasingsequence
1(n+1) =
nn+1 ·
1n + 2n+1
≤ 1n
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Example:
an = (−1)n 1n , n = 1, 2, . . .
yis neithermonotone
decreasing normonotoneincreasing
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Example:
an = (−1)n 1n , n = 1, 2, . . .
yis neithermonotone
decreasing normonotoneincreasing
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Example:
an = (−1)n 1n , n = 1, 2, . . .
yis neithermonotone
decreasing normonotoneincreasing
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Concept: Alternating sequences
A sequence is called alternating iff:
an 6= an+1, n = 1, 2, . . .
and: an · an+1 < 0
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Example: Alternating sequence
an = (−1)n 1n , n = 1, 2, . . .
yis an
alternatingsequence
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Example: Alternating sequence
an = (−1)n 1n , n = 1, 2, . . .
yis an
alternatingsequence
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Example: Alternating sequence
an = (−1)n 1n , n = 1, 2, . . .
yis an
alternatingsequence
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Concept: Convergent/divergent sequences
• limn→∞ an exists
⇔ (an)n is convergent
• limn→∞ an does not exist
⇔ (an)n is divergent
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Concept: Convergent/divergent sequences
• limn→∞ an exists
⇔ (an)n is convergent
• limn→∞ an does not exist
⇔ (an)n is divergent
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Concept: Convergent/divergent sequences
• limn→∞ an exists
⇔ (an)n is convergent
• limn→∞ an does not exist
⇔ (an)n is divergent
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Definition: limn→∞
(an) = A
For any ε > 0 there is an integer Nε
such that for all n ≥ Nε:
|an − A| ≤ ε
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Definition: limn→∞
(an) = A
For any ε > 0 there is an integer Nε
such that for all n ≥ Nε:
|an − A| ≤ ε
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Figure 1: Illustration of the definition of the limit
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Example: A convergent sequence
an = (−1)n 1n , n = 1, 2, . . .
ylimitexists
lim(an) = 0
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Example: A convergent sequence
an = (−1)n 1n , n = 1, 2, . . .
ylimitexists
lim(an) = 0
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Example: A convergent sequence
an = (−1)n 1n , n = 1, 2, . . .
ylimitexists
lim(an) = 0
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Example: A divergent sequence
an = n2, n = 1, 2, . . .
↓
limit doesnot exist
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Example: A divergent sequence
an = n2, n = 1, 2, . . .
↓
limit doesnot exist
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Example: A divergent sequence
an = n2, n = 1, 2, . . .
↓
limit doesnot exist
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A convergence theorem
Every bounded and monotoneincreasing sequence does converge
Every bounded and monotonedecreasing sequence does converge
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A convergence theorem
Every bounded and monotoneincreasing sequence does converge
Every bounded and monotonedecreasing sequence does converge
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A convergence theorem
Every bounded and monotoneincreasing sequence does converge
Every bounded and monotonedecreasing sequence does converge
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Example: bounded and monotone increasing
an = 1− 1n , n = 1, 2, . . .
ylimitexists
limn→∞
(an) = 1
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Example: bounded and monotone increasing
an = 1− 1n , n = 1, 2, . . .
ylimitexists
limn→∞
(an) = 1
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Example: bounded and monotone increasing
an = 1− 1n , n = 1, 2, . . .
ylimitexists
limn→∞
(an) = 1
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2 Series
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Part 2.1
Definition: Series
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Definition: A finite series
A finite series is a
finite sum of numbers.
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Definition: Infinite series
An infinite series is the limit of the sequence ofpartial sums associated with a given sequence(ai)i≥1.
∞∑i=1
ai := limn→∞
{n∑
i=1
ai
}= lim
n→∞{sn}
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Sequence of partial sums:
Given: A sequence (ai)i≥1:
a1 → s1 = a1
a1 a2 → s2 = a1 + a2
a1 a2 a3 → s3 = a1 + a2 + a3
...a1 a2 a3 · · · an → sn = a1 + a2 + a3 + · · · + an
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Sequence of partial sums:
Given: A sequence (ai)i≥1:
a1 → s1 = a1
a1 a2 → s2 = a1 + a2
a1 a2 a3 → s3 = a1 + a2 + a3
...a1 a2 a3 · · · an → sn = a1 + a2 + a3 + · · · + an
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Sequence of partial sums:
Given: A sequence (ai)i≥1:
a1 → s1 = a1
a1 a2 → s2 = a1 + a2
a1 a2 a3 → s3 = a1 + a2 + a3
...a1 a2 a3 · · · an → sn = a1 + a2 + a3 + · · · + an
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Sequence of partial sums:
Given: A sequence (ai)i≥1:
a1 → s1 = a1
a1 a2 → s2 = a1 + a2
a1 a2 a3 → s3 = a1 + a2 + a3
...a1 a2 a3 · · · an → sn = a1 + a2 + a3 + · · · + an
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Sequence of partial sums:
Given: A sequence (ai)i≥1:
a1 → s1 = a1
a1 a2 → s2 = a1 + a2
a1 a2 a3 → s3 = a1 + a2 + a3
...a1 a2 a3 · · · an → sn = a1 + a2 + a3 + · · · + an
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Sequence of partial sums:
i.e., the sequence (ai)i≥1 generates, by theoperation of summation, the partial sums (sn)n≥1:
sn := a1+a2+...+an =n∑
i=1
ai
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Sequence of partial sums:
i.e., the sequence (ai)i≥1 generates, by theoperation of summation, the partial sums (sn)n≥1:
sn := a1+a2+...+an =n∑
i=1
ai
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Notation:
∞∑i=1
ai = limn→∞{sn} = lim
n→∞
{n∑
i=1
ai
}
If this limit exists, i.e. is a finite number,
we call∑∞
i=1 ai a convergent series which isassociated with the sequence (ai)i≥1
.
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Notation:
∞∑i=1
ai = limn→∞{sn} = lim
n→∞
{n∑
i=1
ai
}
If this limit exists, i.e. is a finite number,
we call∑∞
i=1 ai a convergent series which isassociated with the sequence (ai)i≥1
.
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Notation:
∞∑i=1
ai = limn→∞{sn} = lim
n→∞
{n∑
i=1
ai
}
If this limit does not exist we call the
series to be divergent.
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Example 1:
The given sequence (ai )i≥1:an = 1
n2 , n = 1, 2, 3, ...
Question 1: Does∑∞
n=1 an exist ???
1 + 14 + 1
9 + ... =∑∞
n=11n2 converges
Question 2: Find the value of∑∞
n=1 an ?
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Example 1:
The given sequence (ai )i≥1:an = 1
n2 , n = 1, 2, 3, ...
Question 1: Does∑∞
n=1 an exist ???
1 + 14 + 1
9 + ... =∑∞
n=11n2 converges
Question 2: Find the value of∑∞
n=1 an ?
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Example 1:
The given sequence (ai )i≥1:an = 1
n2 , n = 1, 2, 3, ...
Question 1: Does∑∞
n=1 an exist ???
1 + 14 + 1
9 + ... =∑∞
n=11n2 converges
Question 2: Find the value of∑∞
n=1 an ?
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Example 1:
The given sequence (ai )i≥1:an = 1
n2 , n = 1, 2, 3, ...
Question 1: Does∑∞
n=1 an exist ???
1 + 14 + 1
9 + ... =∑∞
n=11n2 converges
Question 2: Find the value of∑∞
n=1 an ?
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Example 1:
Graph of the sequence of partial sums∑n
i=11i2
n=1...30
Prof. Kurt Helmes Sequences and Series
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Example 2:
The given sequence (hn)n≥1:
hn = 1n , n = 1, 2, 3, ...
1 +1
2+
1
3+
1
4... =
∞∑n=1
1
nharmonic series
→ diverges
Prof. Kurt Helmes Sequences and Series
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Example 2:
The given sequence (hn)n≥1:
hn = 1n , n = 1, 2, 3, ...
1 +1
2+
1
3+
1
4... =
∞∑n=1
1
nharmonic series
→ diverges
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Example 2:
The given sequence (hn)n≥1:
hn = 1n , n = 1, 2, 3, ...
1 +1
2+
1
3+
1
4... =
∞∑n=1
1
nharmonic series
→ diverges
Prof. Kurt Helmes Sequences and Series
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Example 2:
Graph of the sequence of partial sums∑n
i=11i
n=1...30
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Part 2.2
Arithmetic Series
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Defintion: Arithmetic Series
Given: An arithmetic sequence (ai)i :
∞∑i=1
ai = limn→∞
n∑i=1
ai
yarithmetic series
Prof. Kurt Helmes Sequences and Series
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Defintion: Arithmetic Series
Given: An arithmetic sequence (ai)i :
∞∑i=1
ai = limn→∞
n∑i=1
ai
y
arithmetic series
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→ Example 1:
Gauß as a “schoolboy”
(“little” Gauß)
cf. the sum of all natural numbers from 1 to 100
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Carl Friedrich Gauß (1777-1855)
Carl Friedrich Gauß, wasborn in Brunswick; he was aworking class kid.
He is considered by many thegreatest mathematican of histime (of all times ???).
His contributions include results in mathematics, astronomy,statistic, physics, etc.
Prof. Kurt Helmes Sequences and Series
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Carl Friedrich Gauß (1777-1855)
Carl Friedrich Gauß, wasborn in Brunswick; he was aworking class kid.
He is considered by many thegreatest mathematican of histime (of all times ???).
His contributions include results in mathematics, astronomy,statistic, physics, etc.
Prof. Kurt Helmes Sequences and Series
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Exercise:
1Find the explicit representation of therecursively defined arithmetic se-quence (an)n, wherean+1 − an = 1, a0 = 0.
2 Find the value (a formula) of the n-thcomponent of the corresponding se-quence of partial sums.
Prof. Kurt Helmes Sequences and Series
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Exercise:
1Find the explicit representation of therecursively defined arithmetic se-quence (an)n, wherean+1 − an = 1, a0 = 0.
2 Find the value (a formula) of the n-thcomponent of the corresponding se-quence of partial sums.
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Exercise:
3Decide whether or not∑∞
i=0 ai converges.
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Solution: Question 1
a0 = 0
a1 − a0 = 1 =⇒ a1 = 1
a2 − a1 = 1 =⇒ a2 = 1 + a1 = 2...
...an − an−1 = 1 =⇒ an = · · · = n
i.e.: (an)n≥0 = (0, 1, 2, 3, 4, 5, ...)
Prof. Kurt Helmes Sequences and Series
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Solution: Question 1
a0 = 0
a1 − a0 = 1 =⇒ a1 = 1
a2 − a1 = 1 =⇒ a2 = 1 + a1 = 2
......
an − an−1 = 1 =⇒ an = · · · = n
i.e.: (an)n≥0 = (0, 1, 2, 3, 4, 5, ...)
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Solution: Question 1
a0 = 0
a1 − a0 = 1 =⇒ a1 = 1
a2 − a1 = 1 =⇒ a2 = 1 + a1 = 2...
...an − an−1 = 1 =⇒ an = · · · = n
i.e.: (an)n≥0 = (0, 1, 2, 3, 4, 5, ...)
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Solution: Question 1
a0 = 0
a1 − a0 = 1 =⇒ a1 = 1
a2 − a1 = 1 =⇒ a2 = 1 + a1 = 2...
...an − an−1 = 1 =⇒ an = · · · = n
i.e.: (an)n≥0 = (0, 1, 2, 3, 4, 5, ...)
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Solution: Question 2
sn = 1 + 2 + 3 + 4 + 5 + ...+ n
+
sn = n + n − 1 + n − 2 + ...+ 1
⇓
2sn = (n + 1) + (n + 1) + ...+ (n + 1)︸ ︷︷ ︸n−times
=⇒ sn =1
2n(n+1)
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Solution: Question 2
sn = 1 + 2 + 3 + 4 + 5 + ...+ n
+
sn = n + n − 1 + n − 2 + ...+ 1
⇓
2sn = (n + 1) + (n + 1) + ...+ (n + 1)︸ ︷︷ ︸n−times
=⇒ sn =1
2n(n+1)
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Solution: Question 2
sn = 1 + 2 + 3 + 4 + 5 + ...+ n
+
sn = n + n − 1 + n − 2 + ...+ 1
⇓
2sn = (n + 1) + (n + 1) + ...+ (n + 1)︸ ︷︷ ︸n−times
=⇒ sn =1
2n(n+1)
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Solution: Question 2
sn = 1 + 2 + 3 + 4 + 5 + ...+ n
+
sn = n + n − 1 + n − 2 + ...+ 1
⇓
2sn = (n + 1) + (n + 1) + ...+ (n + 1)︸ ︷︷ ︸n−times
=⇒ sn =1
2n(n+1)
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Solution: Question 3
The limit of this sequence of partial sums does not exist.
The components/elements of the sequence of partial sums do
not stabilize (around a finite value). The sequence (sn)n is
unbounded.
∞∑i=0
ai = limn→∞
n∑i=0
ai
= limn→∞
{1
2n(n+1)
}=∞
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Solution: Question 3
The limit of this sequence of partial sums does not exist.
The components/elements of the sequence of partial sums do
not stabilize (around a finite value). The sequence (sn)n is
unbounded.
∞∑i=0
ai = limn→∞
n∑i=0
ai
= limn→∞
{1
2n(n+1)
}=∞
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Applicationsof
Arithmetic Sequences & Series:
I Linear depreciation of capital goods
I Simple interest calculations
I Annuities
I Inventory problems
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Applicationsof
Arithmetic Sequences & Series:
I Linear depreciation of capital goods
I Simple interest calculations
I Annuities
I Inventory problems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
Arithmetic Sequences & Series:
I Linear depreciation of capital goods
I Simple interest calculations
I Annuities
I Inventory problems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
Arithmetic Sequences & Series:
I Linear depreciation of capital goods
I Simple interest calculations
I Annuities
I Inventory problems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
Arithmetic Sequences & Series:
I Linear depreciation of capital goods
I Simple interest calculations
I Annuities
I Inventory problems
Prof. Kurt Helmes Sequences and Series
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Example: linear depreciation
R0cost/value of the capital good at timen = 0 (brand-new)
Rnvalue (bookvalue) at the end of year n
r constant rate of depreciation
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Example: linear depreciation
R0cost/value of the capital good at timen = 0 (brand-new)
Rnvalue (bookvalue) at the end of year n
r constant rate of depreciation
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Example: linear depreciation
R0cost/value of the capital good at timen = 0 (brand-new)
Rnvalue (bookvalue) at the end of year n
r constant rate of depreciation
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Example: linear depreciation
Bookvalue after the 1st year :
R1 = R0 − r
Bookvalue after the 2nd year :
R2 = R1 − r ⇔ R2 − R1 = d = −r
Bookvalue after nth year :
Rn = R0 − nr
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Example: linear depreciation
Bookvalue after the 1st year :
R1 = R0 − r
Bookvalue after the 2nd year :
R2 = R1 − r ⇔ R2 − R1 = d = −r
Bookvalue after nth year :
Rn = R0 − nr
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Example: linear depreciation
Bookvalue after the 1st year :
R1 = R0 − r
Bookvalue after the 2nd year :
R2 = R1 − r ⇔ R2 − R1 = d = −r
Bookvalue after nth year :
Rn = R0 − nr
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Example: linear depreciation
Find r so that the bookvalue after 5 years is zero,
i.e. satisfy the requirement R5 = 0.
Idea: Choose
r =R0
5=
purchasing cost
useful lifetime
Recall: Bookvalue after n years:
Rn = R0 − nr
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Example: linear depreciation
Find r so that the bookvalue after 5 years is zero,
i.e. satisfy the requirement R5 = 0.Idea: Choose
r =R0
5=
purchasing cost
useful lifetime
Recall: Bookvalue after n years:
Rn = R0 − nr
Prof. Kurt Helmes Sequences and Series
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Example: linear depreciation
Find r so that the bookvalue after 5 years is zero,
i.e. satisfy the requirement R5 = 0.Idea: Choose
r =R0
5=
purchasing cost
useful lifetime
Recall: Bookvalue after n years:
Rn = R0 − nr
Prof. Kurt Helmes Sequences and Series
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Part 2.3
Geometric Series
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Defintion: Geometric Series
Given: A geometric series (ai)i≥0, i.e. ai = a0 qi , q ∈ R:
∞∑i=0
ai = limn→∞
n∑i=0
ai
= a0
∞∑i=0
qi
ygeometric series
Prof. Kurt Helmes Sequences and Series
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Defintion: Geometric Series
Given: A geometric series (ai)i≥0, i.e. ai = a0 qi , q ∈ R:
∞∑i=0
ai = limn→∞
n∑i=0
ai
= a0
∞∑i=0
qi
ygeometric series
Prof. Kurt Helmes Sequences and Series
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Defintion: Geometric Series
Given: A geometric series (ai)i≥0, i.e. ai = a0 qi , q ∈ R:
∞∑i=0
ai = limn→∞
n∑i=0
ai
= a0
∞∑i=0
qi
ygeometric series
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Example 2
“Big” Gauß
cf. compound interest and annuities
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Story: Part 1
Imagine that at the time when Christ was born
the roman emperor Augustus had been able toinvest
$ 1.23
in a bank account and had been guaranteed an
annual interest rate of 3%; assume interest
payments to be compounded every year.
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Question: Part 1
What was the balance account at
the end of the first year of the new
millenium, i.e. after 2000 years
of compounded interest payments ?
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Solution: Balance account
Initial amount: a0 = 1.23
After 1 year:
a1 =
1 +p
100
a0 = qa0, where q = 1.03
After 2 years: a2 = qa1 = a0q2
...
After n years: an = qan−1 = a0qn
and n = 2000
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Solution: Balance account
Initial amount: a0 = 1.23
After 1 year:
a1 =
1 +p
100
a0 = qa0, where q = 1.03
After 2 years: a2 = qa1 = a0q2
...
After n years: an = qan−1 = a0qn
and n = 2000
Prof. Kurt Helmes Sequences and Series
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Solution: Balance account
Initial amount: a0 = 1.23
After 1 year:
a1 =
1 +p
100
a0 = qa0, where q = 1.03
After 2 years: a2 = qa1 = a0q2
...
After n years: an = qan−1 = a0qn
and n = 2000
Prof. Kurt Helmes Sequences and Series
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Solution: Balance account
Initial amount: a0 = 1.23
After 1 year:
a1 =
1 +p
100
a0 = qa0, where q = 1.03
After 2 years: a2 = qa1 = a0q2
...
After n years: an = qan−1 = a0qn
and n = 2000
Prof. Kurt Helmes Sequences and Series
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Solution: Balance account
After 2000 years:
≈ $ 5.8123 · 1025
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Solution: Balance account
After 2000 years:
$ 58, 123, 869, 869, 669, 184, 628, 080, 369.86
... approx. $ 58 septillions
Prof. Kurt Helmes Sequences and Series
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Balance account over the years n = 1, . . . , 2000
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Balance account over the years n = 333, . . . , 433
Prof. Kurt Helmes Sequences and Series
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Balance account when Columbus discovered Cuba
Prof. Kurt Helmes Sequences and Series
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Story: Part 2
Assume that besides the initial deposit
“relatives” of the emperor had since then deposited$ 1.23
in that very account at the beginning of each newyear.
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Question: Part 2
What was the balance of the account
on December 31, 2000 ?
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Solution: Part 2
The total value of all deposits together with their compoundedinterest is given by (n=2000):
sn = ((deposit on 01.01.2000) + its interest)+ ((deposit on 01.01.1999) + its compound interest)+ ((deposit on 01.01.1998) + its compound interest)...
+ ((deposit when Christ was born) + its compound interest)
Prof. Kurt Helmes Sequences and Series
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Solution:
The balance sn ,n=2000, after 2000 deposits and(compounded) interest payments:
sn = qa0 + q2a0 + · · ·+ a0q2000
= q a0(1 + q + q2 + · · ·+ q1999)
= q a0
1999∑i=0
qi
Prof. Kurt Helmes Sequences and Series
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Problem:
∑1999i=0 qi =: Qk =???, k = 1999
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Solution:
Qk = 1 + q + q2 + q3 + ...+ qk
−qQk = q + q2 + q3 + ...+ qk + qk+1
⇓
Qk − qQk = 1− qk+1 =⇒ (1− q)Qk = 1− qk+1
Prof. Kurt Helmes Sequences and Series
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Solution:
Qk = 1 + q + q2 + q3 + ...+ qk
−qQk = q + q2 + q3 + ...+ qk + qk+1
⇓
Qk − qQk = 1− qk+1 =⇒ (1− q)Qk = 1− qk+1
Prof. Kurt Helmes Sequences and Series
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Solution:
Qk = 1 + q + q2 + q3 + ...+ qk
−qQk = q + q2 + q3 + ...+ qk + qk+1
⇓
Qk − qQk = 1− qk+1
=⇒ (1− q)Qk = 1− qk+1
Prof. Kurt Helmes Sequences and Series
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Solution:
Qk = 1 + q + q2 + q3 + ...+ qk
−qQk = q + q2 + q3 + ...+ qk + qk+1
⇓
Qk − qQk = 1− qk+1 =⇒ (1− q)Qk = 1− qk+1
Prof. Kurt Helmes Sequences and Series
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Solution:
Qk − qQk = 1− qk+1 ⇒ (1− q)Qk = 1− qk+1
⇓
Qk =1− qk+1
1− q=
qk+1 − 1
q − 1if q 6= 1
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Solution:
Qk − qQk = 1− qk+1 ⇒ (1− q)Qk = 1− qk+1
⇓
Qk =1− qk+1
1− q=
qk+1 − 1
q − 1if q 6= 1
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Solution:
Qk =1− qk+1
1− q=
qk+1 − 1
q − 1if q 6= 1
⇓for the special parameter values
Qk =1999∑i=0
qi =(1.03)2000 − 1
(1.03)− 1=
100
3((1.03)2000 − 1)
Prof. Kurt Helmes Sequences and Series
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Solution:
Qk =1− qk+1
1− q=
qk+1 − 1
q − 1if q 6= 1
⇓for the special parameter values
Qk =1999∑i=0
qi =(1.03)2000 − 1
(1.03)− 1=
100
3((1.03)2000 − 1)
Prof. Kurt Helmes Sequences and Series
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Solution:
The solution of the 2nd part of the problem is given by:
sn = q a0 Q1999, n = 2000
and:
q a0 Q1999 = q ao
1999∑i=0
qi ≈ 1.99559 · 1027
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Solution:
The solution of the 2nd part of the problem is given by:
sn = q a0 Q1999, n = 2000
and:
q a0 Q1999 = q ao
1999∑i=0
qi ≈ 1.99559 · 1027
Prof. Kurt Helmes Sequences and Series
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Applicationsof
geometric Sequences & Series:
I geometric depreciation
I compound interest calculations
I annuities
I production theory
I dynamical systems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
geometric Sequences & Series:
I geometric depreciation
I compound interest calculations
I annuities
I production theory
I dynamical systems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
geometric Sequences & Series:
I geometric depreciation
I compound interest calculations
I annuities
I production theory
I dynamical systems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
geometric Sequences & Series:
I geometric depreciation
I compound interest calculations
I annuities
I production theory
I dynamical systems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
geometric Sequences & Series:
I geometric depreciation
I compound interest calculations
I annuities
I production theory
I dynamical systems
Prof. Kurt Helmes Sequences and Series
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Applicationsof
geometric Sequences & Series:
I geometric depreciation
I compound interest calculations
I annuities
I production theory
I dynamical systems
Prof. Kurt Helmes Sequences and Series
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Example: Geometric depreciation
Decreasing amounts of depreciationfor using of a capital good;
the amounts are a fixed percentage ofthe remaining value
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Example: Geometric depreciation
Decreasing amounts of depreciationfor using of a capital good;
the amounts are a fixed percentage ofthe remaining value
Prof. Kurt Helmes Sequences and Series
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Examples: Geometric Depreciation
National accounting rules, f.i. the rate ofdepreciation satisfies:
• p%≤ 200
lifetime%
and
• p%≤ 20%
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Examples: Geometric Depreciation
National accounting rules, f.i. the rate ofdepreciation satisfies:
• p%≤ 200
lifetime%
and
• p%≤ 20%
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Example: Geometric Depreciation
Formula:
R0 = initial value (purchasing price)
A1 =p
100R0 = 1st amount of depreciation
⇒ R1 = R0 − A1 = R0 −p
100R0 =
1− p
100
R0
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Example: Geometric Depreciation
Formula:
R0 = initial value (purchasing price)
A1 =p
100R0 = 1st amount of depreciation
⇒ R1 = R0 − A1 = R0 −p
100R0 =
1− p
100
R0
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Example: Geometric Depreciation
Formula:
A2 =p
100R1 = 2nd amount of depreciation
R2 = R1 − A2 =
1− p
100
R0 −p
100
1− p
100
R0
=
1− p
100
2
R0 = q2R0,
where q =
1− p
100
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Example: Geometric Depreciation
Formula:
A2 =p
100R1 = 2nd amount of depreciation
R2 = R1 − A2 =
1− p
100
R0 −p
100
1− p
100
R0
=
1− p
100
2
R0 = q2R0,
where q =
1− p
100
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Example: Geometric Depreciation
Formula:
A2 =p
100R1 = 2nd amount of depreciation
R2 = R1 − A2 =
1− p
100
R0 −p
100
1− p
100
R0
=
1− p
100
2
R0 = q2R0,
where q =
1− p
100
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Example: Geometric Depreciation
Formula:
An =p
100Rn−1 = nth amount of depreciation
Bookvalue at the end of the nth year:
Rn = qnR0
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Example: Geometric Depreciation
Formula:
An =p
100Rn−1 = nth amount of depreciation
Bookvalue at the end of the nth year:
Rn = qnR0
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Example: Geometric Depreciation
Table:
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Part 2.4
Some Properties
of Series
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Criteria of convergence:
Condition on q so that
geometric series
∞∑i=0
qi
does converge.
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Criteria of convergence:
Condition on q so that
geometric series
∞∑i=0
qi
does converge.
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Criteria of convergence:
A simple idea:Let q 6= 1, then
∞∑i=0
qi = limn→∞
{n∑
i=0
qi
}= lim
n→∞
{1− qn+1
1− q
}
=1
1− q− lim
n→∞qn+1
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Criteria of convergence:
A simple idea:Let q 6= 1, then
∞∑i=0
qi = limn→∞
{n∑
i=0
qi
}= lim
n→∞
{1− qn+1
1− q
}
=1
1− q− lim
n→∞qn+1
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Criteria of convergence:
Hence,
∑∞i=0 qi =
1
1− qconverges if |q| < 1
∑∞i=0 qi diverges if |q| ≥ 1
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Criteria of convergence:
(a special case of the dominating principle)
Assumption:
(ai )i≥0 is a sequence such that:
|ai | ≤ qi fur 0 < q < 1 und i ≥ i0
Claim:
∞∑i=0
ai is a convergent series
Prof. Kurt Helmes Sequences and Series
![Page 192: Lectures 1 & 2 : Sequences and Series...Lectures 1 & 2 : Sequences and Series Prof. Kurt Helmes Institute of Operations Research Humboldt-University of Berlin 6th Summer-School Havanna](https://reader034.fdocuments.in/reader034/viewer/2022042915/5f526bec8029d24d00653c71/html5/thumbnails/192.jpg)
Criteria of convergence:
(a special case of the dominating principle)
Assumption:
(ai )i≥0 is a sequence such that:
|ai | ≤ qi fur 0 < q < 1 und i ≥ i0
Claim:
∞∑i=0
ai is a convergent series
Prof. Kurt Helmes Sequences and Series
![Page 193: Lectures 1 & 2 : Sequences and Series...Lectures 1 & 2 : Sequences and Series Prof. Kurt Helmes Institute of Operations Research Humboldt-University of Berlin 6th Summer-School Havanna](https://reader034.fdocuments.in/reader034/viewer/2022042915/5f526bec8029d24d00653c71/html5/thumbnails/193.jpg)
Criteria of convergence:
(a special case of the dominating principle)
Assumption:
(ai )i≥0 is a sequence such that:
|ai | ≤ qi fur 0 < q < 1 und i ≥ i0
Claim:
∞∑i=0
ai is a convergent series
Prof. Kurt Helmes Sequences and Series
![Page 194: Lectures 1 & 2 : Sequences and Series...Lectures 1 & 2 : Sequences and Series Prof. Kurt Helmes Institute of Operations Research Humboldt-University of Berlin 6th Summer-School Havanna](https://reader034.fdocuments.in/reader034/viewer/2022042915/5f526bec8029d24d00653c71/html5/thumbnails/194.jpg)
Example of the criterium:
Let ai =i
2i; the series
∞∑i=0
i
2iconverges
Proof:i
2i≤
3
4
i
, if i ≥ 1, i.e. q =3
4and i0 = 1
Prof. Kurt Helmes Sequences and Series
![Page 195: Lectures 1 & 2 : Sequences and Series...Lectures 1 & 2 : Sequences and Series Prof. Kurt Helmes Institute of Operations Research Humboldt-University of Berlin 6th Summer-School Havanna](https://reader034.fdocuments.in/reader034/viewer/2022042915/5f526bec8029d24d00653c71/html5/thumbnails/195.jpg)
Example of the criterium:
Let ai =i
2i; the series
∞∑i=0
i
2iconverges
Proof:i
2i≤
3
4
i
, if i ≥ 1, i.e. q =3
4and i0 = 1
Prof. Kurt Helmes Sequences and Series
![Page 196: Lectures 1 & 2 : Sequences and Series...Lectures 1 & 2 : Sequences and Series Prof. Kurt Helmes Institute of Operations Research Humboldt-University of Berlin 6th Summer-School Havanna](https://reader034.fdocuments.in/reader034/viewer/2022042915/5f526bec8029d24d00653c71/html5/thumbnails/196.jpg)
Finally!!! ;)
The End
Prof. Kurt Helmes Sequences and Series