5.7. Sequences and series - Poenitz

5.7. Sequences and series
Natural processes like growth of populations of particles, cells or animals, but also share values, prices, exchange rates, sales
numbers etc. evolve in many little steps. Every single chemical reaction, cell division, birth, death, sale and buy action
produces a small jump of the population or rate. In a mathematical context, these steps are discontinuities incommensurable
with the continuous functions (see 4.7. and 5.6.) usually used to model these processes. Instead of continuous functions
discrete sequences and series provide more realistic simulations of these processes.
5.7.1. Recursive and general formulae for sequences
Example: Exercises on sequences and series No. 1
25
Definition
A function with domain D = ℕ = {0; 1; 2; ...} is called a
sequence. Instead of the variable x we use the index n; the
value f(x) of f at x becomes the nth term an.
y
20
an
f(x)
1
2
15
Example:
10
f(x) = 2x + 3 with domain D = ℝ is a function
5
an = 2n + 3 with domain D = ℕ is a sequence
0
Exercises on sequences and series No. 2
x resp. n
0
3
4
5
6
7
8
9
10
Definition
1. A general formula gives the value of an depending only on n. As in a function f(x) the general formula makes it possible
to find the value of an simply by plugging in n without having to know the value of the previous terms.
2. A recursive formula gives the value of an + 1 depending on the previous term a n. Because an depends on an − 1 and so on
actually all previous term have to be known. On the other hand the recursive formula often is much simpler and easier to
understand than the general formula which may even be unknown!
Example: Recursive and general formula for exponential growth (see 4.7.4.)
An amount of a0 = 1000 € is deposited in a savings account with a compound interest of 15 %. Give a recursive formula and a
general formula for the amount after n years.
Solution:
recursive: an+1 =∙an + 0,15 an = 1,15∙an with a0 = 1000
general: an = 1,15∙an−1 = 1,152∙an−2 = ... = 1,15n∙a0 = 1,15n∙1000.
Examples for sequences used as approximation for irrational numbers in analysis:
1
a
1. recursive approximation of a squareroot a = lim xn with xn + 1 =
(see 5.7.5)
xn
n 
2
xn
2
2. recursive approximation of π = lim xn with π2n = n 2 2 1
n
n 
(see 5.7.6 and 5.7.7)
n
3. general approximation of Euler’s number e = lim en with en = 1
n 
1
n
n
(see 5.7.8)
4. recursive approximation of a zero (Newton-Raphson-method) x = lim xn with xn + 1 = xn −
n 
an
Arithmetic sequences
A sequence with
 the recursive formula an + 1 = d + an
 the general formula an = n‧d + a0
describes a linear growth with
 initial value a0
 common difference d
and is called a arithmetic sequence.
an + 1
d
an
a2
d
a1
f (x n )
(see 5.3.6)
f '(x n )
n∙d
d
a0
0
1
2
n
n+1
n
1
Geometric sequences
A sequence with
 the recursive formula an + 1 = 1
an
p
∙an
100
an + 1
n
p
∙a0
100
describes an exponential growth with
 initial value a0
 percentage p
p
 common ratio q = 1
100
and is called a geometric sequence.
 the general formula an = 1
an
a1
a0
0
1
a2
2
n
n+1
n
Exercises on sequences and series No. 3
Example for converting a general formula into a recursive formula
Fin the recursive formula of an = n3 − 3n2.
Solution:
an+1 − an = [(n + 1)3 − 3(n + 1)2] − [n3 − 3n2]
= n3 + 3n2 + 3n + 1 − 3n2 − 6n − 3 − n3 + 3n2
= 3n2 − 3n − 2
⇒ an+1 = an + 3n2 − 3n − 2 mit a0 = 0
Exercises on sequences and series No. 4
A formula for the first n natural numbers = summation rule of the arithmetic series (see 5.7.4.) with an = n:
1
1 + 2 + 3 + ... + (n − 2) + (n − 1) + n = (n + 1) + (n + 1) + (n + 1) + ... = n∙(n + 1), because
2
by pairing numbers with equal sums you get
or
1
n pairs with sum (n + 1), if n is even
2
1
1
(n – 1) pairs with sum (n + 1) and the lone value (n + 1) in the middle, if n is odd
2
2
Example for the conversion of a recursive formula into a general formula
Find a general formula for the sequence an+1 = an + 4n − 2 with a0 = 1.
Solution:
an = an−1 + 4(n − 1) − 2
= an−1 + 4n − 6
= an−2 + 4(n − 1) + 4n − 6 − 6
= an−3 + 4(n − 2) + 4(n − 1) + 4n − 6 − 6 − 6
.
.
.
1
= a0 + 4(1 + 2 + 3 + ... + n) − n∙6 = 4∙ n∙(n + 1) − 6n
2
= 1 + 2n2 + 2n − 6n
= 1 + 2n2 − 4n
= 2n(n − 2) + 1
∣ apply summation rule from above
Exercises on sequences and series No 5
2
5.7.2. Monotone and bounded sequences (see 5.3.1.)
Definition
A sequence (an) is called

monotonically increasing, if an+1 ≥ an ⇔ an+1 − an ≥ 0 (difference criterion)
a
for positive terms an and an + 1 the condition n 1 ≥ 1 is equivalent (ratio criterion)
an

monotonically decreasing, if an+1 ≤ an ⇔ an+1 − an ≤ 0 ⇔
an
an
1
≤ 1 viz for positive an and an+1 we have
an
1
an
≤ 1.
for all n ∈ ℕ. The ratio criterion is convenient in case of an being a product.
Example 1:
Examine the sequence an = n3 − 16n on monotonic behaviour
Solution
Since an is a sum the difference criterion is suitable an + 1 – an ≥ 0
an+1 − an = n3 + 3n2 + 3n + 1 − 16n − 16 − n3 + 16n
= 3n2 + 3n −15
= 3(n2 + n − 5)
> 0 für n ≥ 2
⇒ an (strictly) monotonically increases for n ≥ 2.
Example 2:
Examine the sequence an = n4∙2−n on monotonic behaviour
Solution
Since an is a product and always positive the ratio criterion is convenient:
an
an
1
=
(n 1)4 2 (n
n4 2 n
(n 1)4
n4 2
1 n 1
=
2
n
an
an
1
≥1
1)
=
=
4
4
1
1
2
1
.
n
→ 1 for n → ∞
→
1
1
∙1 = for n → ∞
2
2
With the help of the GDC in sequential mode we quickly calculate some values:
a6
a
≈ 1,04 > 1 und 7 ≈ 0,93 < 1
a5
a6
⇒ an is (strictly) monotonically decreasing for n ≥ 6.
Exercises on sequences and series No. 6
Definition
A sequence (an) is
 bounded above, if an ≤ S
 bounded below, if an ≥ S
for an upper resp. lower bound S ∈ ℝ and all n ∈ ℕ
3
Example:
Examine if the sequence an = n4∙2−n is bounded and give reasons.
25
Solution:
20
an is bounded below,
15
since an > 0 for all n ∈ ℕ.
y
an
f(x)
1
2
10
an is bounded above,
since it decreases after the maximal value a6 ≈ 20,25.
(see example 2 above: the existence of an upper bound can only be
shown if you know that (an) decreases.
5
0
x resp. n
0
3
4
5
6
7
8
9
10
Exercises on sequences and series No. 7
5.7.3. Limit of a sequence for n → ∞ (see also 5.1.1.)
Definition
A sequence (an) converges or tends to the limit a, if for any (especially small) ε > 0 existsts a nε, so that all terms after nε
deviate by less than ε from this limit: ∣an − a∣ < ε for n ≥ nε. The notation is an → a für n → ∞ or lim an = a
n
Example
Find the number nε so that all terms of the sequence (an) = (−2)−n + 1
deviate by less than ε = 0,1 resp. 0,01 resp. 0,001 from the limit a = 1?
an
Solution:
∣an − a∣ < ε
∣(−2)−n ∣ < ε
2−n < ε
−n ln(2) < ln(ε)
ln( )
n >−
ln(2)
2
2ε
1
n
⇒ n0,1 = 4, n0,01 = 7 and n0,001 = 10
1
5
Exercises on sequences and series No. 8
Theorem: If a sequence is monotonic and bounded it tends to a limit.
an
Proof:
S=2
Assume (an) is monotonically increasing and has the upper
limit S.
I3
Then we can find a least upper bound s with the following
I2
a=1
decreasing sequence of intervals (In):
I1
We start with I0 = [a0;S] and choose the next interval
I0
recursively as follows : Ik = upper half of Ik − 1, if this upper half
n
contains terms of (an) and Ik = lower half of Ik – 1, if the upper
1
5
half is empty.
Because every interval Ik is only half as large as and included in the previous interval Ik − 1 the upper and lower bounds of this
telescoping chain (Ik) of intervals converge to the same limit s. s is an upper bound of (an), because all upper bounds of Ik are
an upper bound of (an). s is also the least upper bound because each interval In contains elements of (an). s is also the limit of
(an) for n → ∞ because each interval I k contains s and all elements of (an) beyond a certain n(k). But since the diameter of Ik is
S∙2−k the deviation |s – an| < S∙2−k for all n > n(k) and for any given ε you can find a k with S∙2−k < ε.
4
Example:
Show that the sequence an =
n 2
is monotonic and bounded and therefore must converge.
2n 1
Solution:
Monotonicity with the ratio criterion:
an 1
(n 1) 2 2n 1
=
∙
an
2(n 1) 1 n 2
=
(n 1)(2 n 1)
??
(n 2)(2n 3)
→ Better try difference criterion:
(n 1) 2
n 2
an+1 − an =
−
2(n 1) 1 2n 1
n 2
n 1
−
2n 1
2n 3
(n 1)(2n 1) (n 2)(2n
=
(2n 1)(2n 3)
=
=
=
2n 2
3)
n 1 (2 n 2 n 6)
(2n 1)(2n 3)
5
(2n 1)(2n
3)
> 0 for all n ∈ ℕ
⇒ an is (strictly) monotonically increasing.
An upper boundary is for example S = 1, because
an < 1
| plug in
n 2
<1
| ∙(2n + 1) > 0
2n 1
n− 2 < 2n + 1 | − n ; − 1
−3 < n
which holds for all n ∈ ℕ.
Since (an) is increasing and has an upper bound, it must converge to a limit.
This limit can be found easily by polynomial long division as in rational functions
n 2
1
5
1
= lim
= .
lim
n
n
2 4n 2
2n 1
2
→ 0 for n → ∞
Exercises on sequences and series No. 9
5
5.7.4. Series
Sigma notation and series
The corresponding series of a given sequence (an) is the sequence of
an
n
a2
a3
a4
a k := a0 + a1 + … + an,.
the sums of the first n terms
k 0
The letter Sigma Σ symbolizes the sum as does the S in an integral.
n
a1
a k corresponds
The summation of a sequence (an) to form a series
a5
k 0
n 1
to the integration of a function f(x) to its Integral
f (x)dx .
a0
0
n
ak =
The limit of a converging series is denoted lim
n
k 0
ak .
0
1
2
3
4
n bzw. x
5
k 0
n
Likewise the limit of a converging integral is denoted lim
f (x)dx =
n
0
f (x)dx and called improper integral.
0
(The use of infinity ∞ as an upper bound actually is a contradiction and therefore an improper use of notation!)
Exercises on sequences and series No. 10
The arithmetic series
The summation of the arithmetic sequence an = a0 + d∙n results in the arithmetic series
n
a k = a0 +
a1
+
a2
+…+
an−1
+
an
k 0
= a0 + (a0 + 1∙d) + (a0 + 2∙d) + … + (a0 + (n – 1)∙d) + (a0 + n∙d)
= (n + 1)∙a0 + 1∙d +
2∙d + … +
= (n + 1)∙a0 + (n + 1)∙d + (n + 1)∙d + …
1
n∙(n + 1)d
2
Integrating the corresponding linear function
f(x) = a0 + dx leads to the integral
= (n + 1)∙a0 +
n 1
f (x)dx = a 0 x
0
1 2
dx
2
(n – 1)∙d +
n∙d
1
n∙(n + 1)∙d, if n even
2
1
1
1
(n – 1)(n + 1)d + (n + 1)∙d = n∙(n + 1)∙d, if n odd
2
2
2
an
an+1
an
a2
n 1
a1
0
a0
1
(n + 1)2d.
2
0
The integral is larger than the corresponding sum, since the function
is monotonically increasing and passes above the rectangles of the sum.
= (n + 1)a0 +
1
2
n
n+1
n
Example:
4
9
a k = 12 and
Determine the initial value a0 and the common difference d of the arithmetic sequence (an) with
k 0
a k = 29.
k 0
Solution:
4
a k = 5a0 + 10d = 12 ⇔ 10a0 + 20d = 24
With
k 0
9
a k = 10a0 + 45d = 29
and
k 0
subtracting the two equations results in 25d = 5 ⇒ common difference d =
12
1
and initial value a0 =
− 2d = 2
5
5
Exercises on sequences and series No. 11
6
The geometric series
The summation of the geometric sequence an = a0∙qn results in the geometric series
n
a k = a0 + a 1 +
a2
+ … + an
= a0 + q‧a0 + q ‧a0 + … + q ‧a0
= a0(1 + q + q2 + … + qn)
q n 1  1
= a0‧
, since
q 1
2
q−1
y
k 0
n
ln(q)
−1
(1 + q + q2 + … + qn)‧(q − 1) = q + q2 + q3 + … + qn + 1
−(1 + q + q2 + … + qn)
n+1
=q
–1
x
1
an
an + 1
If counting starts with n = 1 we have
an
n
a k = a1 + a 2 +
a3
+ … + an
k 1
= a1 + q‧a1 + q2‧a1 + … + qn − 1‧a1
= a1(1 + q + q2 + … + qn − 1)
qn 1
= a1‧
q 1
a1
a0
0
1
a2
2
n
n+1
n
The integration of the corresponding exponential function f(x) = a0∙qx results in the integral
n 1
0
a qx
f (x)dx = 0
ln(q)
n 1
= a0‧
0
q n 1  1
ln(q)
Becaus q – 1 > ln(q) (see drawing above right) the integral is slightly larger than the corresponding sum, since the function
is monotonically increasing and passes above the rectangles of the sum.
Convergence of geometric series
For decreasing sequences with negative percentages p < 0 resp. common ratios q < 1 the power qn vanishes for n → ∞ and
n
a
1
q n 1  1
a k = lim
a k = lim a0‧
the geometric series converges to
= a0∙
= 0
n
n
q 1 1 q
q 1
k 0
k 0
Example:
2
5
a k = 775 and
Determine the initial value a0 and the common ratio d oft the geometric sequence an with
k 0
a k = 781,2.
k 0
Solution:
2
a k = a0∙
With
k 0
q3 1
= 775 and
q 1
5
a k = a0∙
k 0
q6 1
q6
= 781,2 dividing the two equations results in a0∙ 3
q 1
q
1
781, 2
=
⇔
1
775
(q 3 1)(q 3 1)
= 1,008 ⇒ q3 + 1 = 1,008 ⇔ q3 = 0,008 ⇒ common ratio q = 0,2 and initial value a0 = 625.
q3 1
Exercises on sequences and series No. 12
integral criterion for the boundedness of series
Usually with the help of the various integration rules integrals are much easier to calculate than the corresponding sums. In
many cases a good upper bound can be found by integrating a (possibly slightly adapted) corresponding function in suitable
bounds. The function and the bounds have to be chosen so that the function is always above the rectangles of the sum inside
the bounds. Likewise a lower bound may be found by integration a suitable function passing inside the rectangles of the sum
Example for the use of the integral criterion
n
1
1
1
1
1
Show that the series
= 3 + 3 + 3 + ... + 3 is increasing and bounded above and therefore must converge.
3
3
1
2
n
k 1 k
7
Solution
Let an =
n
1
n
a k the corresponding series.
be the generating sequence and sn =
3
k 1
The series sn is monotonically increasing, since sn + 1 − sn = an + 1 =
1
(n
1)3
The generating sequence an is monotonically decreasing, since an + 1 − an =
sn is bounded above, since
1
1
1
1
sn = 3 + 3 + 3 + ...+ 3
3
1
2
n
n
<1+
1
1
x3
=1−
=3−
x2
2
n2
2
1
1)3
(n
−
1
n3
=
n3
(n 1)3
< 0 for all n ∈ ℕ.
(n 1)3 n 3
y
dx (integral criterion, attention: adapted bounds!)
2
=1+
> 0 for all n ∈ ℕ
n
1
+2
1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
0
1
2
3
4
5
6x
n2
< 3 for all n ∈ ℕ.
1
< 3.
3
k 1 k
However the exact value of this limit is still unknown! Leonhard Euler showed in 1736 that
2
2
1
1
1
1
1
1
1
1
=
+
+
+
...
=
and
=
+
+
+
...=
2
1) 2
6
8
13 23 33
13 33 53
k 1 k
k 0 (2k
Therefore sn must have a limit lim sn =
n
Exercises on sequences and series No. 13
5.7.5. Mathematical induction
Proof by mathematical induction (recursive proof)
In order to prove a statement for all n ∈ ℕ it suffices to show
1. that the statement holds for n = 0 (base case, starting point)
2. that if the statement hold for n then it also holds for n + 1. (inductive
step, chain reaction)
Example: (summation formula for the first n natural numbers)
1
Theorem: For all n ∈ ℕ we have 1 + 2 + 3 + ... + n = n(n + 1)
2
Proof:
1
base case n = 1: 1 = ∙ ∙1∙(1 + 1) = 1
2
inductive step n ⇒ n + 1:
inductive hypothesis for some n: 1 + 2 + 3 + ... + n =
1
n(n + 1)
2
1
(n + 1)(n + 2)
2
1
1
3
Plug the inductive hypothesis into the left side: 1 + 2 + 3 + ... + n + n + 1 = n(n + 1) + n + 1 = n2 + n + 1
2
2
2
1
1 2 3
right side: (n + 1)(n + 2) = n + n + 1 = left side, qed
2
2
2
we have to show the statement for n + 1: 1 + 2 + 3 + ... + n + n + 1 =
Exercises on sequences and series Nos. 14 und 15
8