Transcript Introduction to Algebra

Geometric Series
Multiplicative
Recursion
Series

What is a series ?

A sum of a sequence of numbers

Definition:
A finite series is a sum of form
a1 + a2 + a3 + ... + an
for some positive integer n
NOTE:
A finite series has last term an
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Series

Finite Series Example
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Sequence is
{ an } =
{
(–1)n–1 4
2n – 1
}
4 –4 4 –4
= 1, 3 , 5, 7 ,
•••
nth partial sums are
n
Sn =
∑
k=1
(–1)k–1 4
2k – 1
4
4
–4
–4 • • • (–1)n–1 4
= 1 + 3 + 5 + 7 +
+
2n – 1
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Series

What is an infinite series ?

A sum of a sequence of numbers

Definition:
An infinite series is a sum of form
a1 + a2 + a3 + ... + an + ...
NOTE:
An infinite series has no last term
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Geometric Sequences
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Definition

A geometric sequence is a function defined
on the set of positive integers of form
f(n) = an = ran–1
where r is the common ratio and a1 is a
constant
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Geometric Sequences

A geometric sequence is a function defined
on the set of positive integers of form
f(n) = an = ran–1
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By induction we can show
r a n–1 = r n–1 a1
* that
Note:
an
an–1 = r , the ratio of successive terms
* Write out a few terms, starting with a
2
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Geometric Sequences
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Partial Sums
Recall that an = r an–1 = r n–1 a1
WHY ?
Sn = a1 + a2 + a3 + ... + an–1 + an
= a1 + ra1 + r2a1 + ... + rn–2a1 + rn–1a1
rSn = ra1 + r2a1 + r3a1 + ... + rn–1a1 + rna1
Sn – rSn = a1 – r n a1
Now, factoring …
Sn (1 – r) = a1 (1 – r n )
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Geometric Sequences

Partial Sums
Sn (1 – r) = a1 (1 – r n )
Thus,
Sn =
a1(
1 – rn
1–r
)
Question:
What happens to Sn as n
It depends on r …
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∞?
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Geometric Series

What Is A Geometric Series ?
Definition:
n
A series ∑ ak such that
k=1
ak+1 = rak
for some constant r and for all k,
k = 1, 2, 3, … , n is a geometric
series with common ratio r
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Geometric Series
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Geometric Sequences and Series
Alternate view:
n
A geometric series ∑ ak with
k=1
common ratio r is the sum of a
geometric sequence { ak } with
common ratio r
ak
Note: a
=r
k–1
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Geometric Series
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Partial Sums
Recall that Sn =
a1(
If r < 1 then r n
Thus
Sn
1 – rn
1–r
)
0 as n
a1
1–r
∞
∞
= ∑ ak = S
k=1
However …
if r  ≥ 1 then S does not exist !!
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Geometric Series
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Examples: Geometric Series
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1. Sequence is { an } = { 21–n }
1 1 1 ••• 1
= 1, 2 , 4 , 8 , , 2n–1
Here r = 2–1 and partial sums are:
S1 = 1
1
S2 = 1 + 2
S3 = 1 + 12 + 14
1
1
1
Sn = 1 + 2 + 4 + 8 +
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Geometric Series
•••
+
1
2n–1
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Geometric Series
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Examples: Geometric Series
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1. Sequence is { an } = { 21–n }
1
1
1
Sn = 1 + 2 + 4 + 8 +
We can show that Sn
Graphically
0
+
1
2n–1
2 as n
1
2
1
•••
∞
1
4
•••
1
2
So, S = 2
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Geometric Series
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Examples: Geometric Series
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2. Sequence is { an } = { 2n–1 }
Note change
in exponent
Here r = 2 and partial sums are:
Sn = 1 + 2 + 4 + 8 +
1 – rn
= a1 1 – r
(
•••
) = –1 + 2
We can show that Sn
+ 2n–1
n
∞ as n
∞
So …
S does not exist !
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Geometric Series

Examples
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3.
a1 = 1
with common ratio 3–1
| r | = 3–1 < 1 so r n
1 – rn
and Sn = a1
1–r
1
3
Then S =
1 =
2
1– 3
)
(
∞
0 as n
a1
1–r
= S
∞
Thus S =
∑ ak
DOES exist !
k=1
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Geometric Series
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Examples
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4.
a1 = 5
with common ratio 0.7
Since | r | = 0.7 < 1
then
S=
Thus
a1
1–r
S=
=
5
1 – 0.7
―
= 16.66
∞
―
∑ ak = 16.66
k=1
Does exist !
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Geometric Series
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Examples
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5.
a1 = 2
with common ratio -1
Note that if we use the formula
a1
2
=
S =
=1
1–r
1 – (-1)
we get S = 1 for the infinite sum !
BUT …
Recall:
an = ran–1 = r n–1 a1
= (-1)n–1 (2)
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Geometric Series
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Examples
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5.
a1 = 2
Thus
with common ratio -1
an = (-1)n–1 (2)
∞
∞
k=1
k=1
∑ ak = ∑ (-1)n–12
=2–2+2–2+2–2+…
So, sequence of partial sums is
{ Sn } = 2, 0, 2, 0, 2, 0, …
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Geometric Series
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Examples
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5.
a1 = 2 with common ratio -1
{ Sn } = 2, 0, 2, 0, 2, 0, …
Partial sums Sn = 2 for n odd ,
but
Sn = 0 for n even
Thus Sn
S for any number S !
That is, the infinite sum does not exist !
What happened here ?
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Geometric Series
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Examples
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5.
a1 = 2
with common ratio -1
So S = 1 BUT S does not exist !
Recall the condition for S =
that is
a1
1–r
,
|r|<1
Does r = -1 satisfy this condition ?
Note:
Applying a formula without satisfying its
conditions yields unpredictable results !
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Geometric Series
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Examples
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6.
a1 = -6
with common ratio 4
1 – rn
Sn = a1 1 – r
(
)
= 2(1 – 4n)
Thus
– ∞ as n
Sn
Thus
∞
∞
S = ∑ ak does not exist !
k=1
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Geometric Series
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Examples
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7.
a1 = -6
with common ratio
1 – rn
Sn = a1 1 – r
(
) = -6(
= -8(1 – (¼)n
Thus
Sn
Thus
-8 as n
∞
¼
S = ∑ ak = -8
1 – (¼)n
1–¼
)
)
∞
does exist !
k=1
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Geometric Series
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Application Example
$1000 is deposited in an account paying 5%
interest annually. After five years, with no
other deposits or withdrawals, how much is
in the account?
Let In = interest accrued in year n
Account balance = 1000 + (I1 + I2 + I3 + I4 + I5)
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Geometric Series
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Application Example
Account balance = 1000 + (I1 + I2 + I3 + I4 + I5)
where
I1 = .05(1000)
I2 = .05(1000) + .05I1 = 1000(.05 + .052)
I3 = .05(1000) + .05I2 = 1000(.05 + .052 + .053)
I4 = .05(1000) + .05I3 = 1000(.05 + .052 + .053 + .054)
I5 = .05(1000) + .05I4 = 1000(.05 + .052 + .053 + .054 + .055)
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Geometric Series
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Application Example - Review
Account balance = 1000 + (I1 + I2 + I3 + I4 + I5)
= 1000 (1 + 5(.05) + 4(.05)2 + 3(.05)3 + 2(.05)4 + (.05)5
(
= 1000
)
5
1 +
)
∑ (5 – (k – 1)) (.05)k
k=1
Note: For n years the balance is
(
1000
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n
1 +
)
∑ (n – (k – 1)) (.05)k
k=1
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Think about it !
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