5.7. Exercises on sequences and series - Poenitz

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5.7. Exercises on sequences and series
Exercise 1: linear and exponential growth
A square with side length 1 dm consecutively grows smaller squares on three of its four
sides. The squares added in the n th step have only one third of the side length of the
squares added in the previous step n – 1
a) Calculate the circumference Un after n = 0, 1, 2, 3 and 4 steps.
b) What is the difference Un+1 − Un of the circumference from step n to step n + 1?
c) Find an expression of Un+1 in terms of Un.
d) Find an expression of Un in terms of n.
e) Calculate the area An of the shape after n = 1, 2, 3 and 4 steps
f) What is the difference An+1 − An of the area from step n to step n + 1?
g) Find an expression of An+1 in terms of An.
h) Find an expression of An in terms of n.
1 xn 1
.
1 x
Calculate A100 and U100. Describe the trend of Un and An for ever larger n. Draw
conclusions about the circumference of natural places like for example an island.
Determine the limit lim An.
Hint: 1 + x + x2 + x3 + ... + xn =
i)
j)
n
Exercise 2: Calculating terms with recursive and general formulae
Calculate the first 5 terms a0, ..., a4 and draw a diagram. Can you guess the limit for n → ∞?
1
a) an = 100∙2−n
e) an+1 = an +
with a0 = 3
2
1
b) an = 100 − 50∙2−n
f) an+1 = an + an with a0 = 3
2
1
1
c) an =
g) an+1 = an + (5 − an) with a0 = 3
2
n 1
1
d) an = (n + 1)(n + 2)
h) an+1 = an +
an∙(5 − an) with a0 = 3
20
Exercise 3: Finding recursive and general formulae for given terms of a sequence
Find the recursive formula and the general formula for the sequence with the terms given:
2
4
1
3
a) a0 = 1; a1 = 3; a2 = 5; a3 = 7; a4 = 9
e) a0 = 0; a1 = ; a2 = ; a3 = ; a4 =
2
4
3
5
2
4
16
8
b) a0 = 3; a1 = 6; a2 = 12; a3 = 24; a4 = 48
f) a0 = 1; a1 = ; a2 = ; a3 =
; a4 =
3
9
81
27
7
11
5
c) a0 = 2; a1 = 6; a2 = 18; a3 = 54; a4 = 162
g) a0 = −1; a1 = 1; a2 = ; a3 =
; a4 =
5
7
3
1
4
1
16
d) a0 = 2; a1 = 5; a2 = 10; a3 = 17; a4 = 26
h) a0 = 0; a1 = ; a2 = ; a3 = ; a4 =
3
9
3
81
Exercise 4: Converting general formulae into recursive formulae
Find a recursive formula for the given general formula
a) an = 3n + 2
b) an = n2 − 2n
c) an = 3−n
d) an =
n
n 1
Exercise 5: Converting recursive formulae into general formulae
Find an general formula for the given recursive formula
a) an+1 = an − 3 with a0 = 2
c) an+1 = an + 2n + 2 with a0 = 0
b) an+1 = 0,8∙an with a0 = 20
d) an+1 = an + 2n + 1 with a0 = 0
1
Exercise 6: Monotonicity of a sequence
Examine the following sequences on monotonic increasing or decreasing behavior:
n 1
a) an =
b) an = n 2 n
c) an = n3 − 3n2
d) an = n2∙2−n
n 1
Exercise 7: Boundedness of a sequence
Examine the following sequences on upper and lower bounds:
n
a) an =
b) an = n 2 n
c) an = n2 − n3
n 1
d) an = n2∙3−n
Exercise 8: Limit of a sequence
Find the limit lim an and give reasons.
n
n 2
a) an =
n 1
b) an =
1
n
n2
c) an =
n
1
sin(n)
n
d) an = n3∙2−n
Exercise 9: Convergence of a sequence
Examine the sequence (an) for n ≥ 1 on monotonicity and boundedness. Then draw a conclusion about its convergence:
a) an =
1 4n
1 2n
b) an =
n 1
2n
c) an =
2n
2n
3n
3n
d) an =
n n 10
n2
Exercise 10: Series and sigma notation
Fill in the blanks:
n 1
n
ak
series
generating sequence an
Integral
corresponding function f(x)
k 0
f (x)dx
1
1
n
1
1
1
+
+
+…
3
9
27
1
1
1
1+
+
+
+…
4
9 16
1+
1
1
1
=
−
x
x 1
x(x 1)
Exercise 11: Arithmetic series
20
100
a k and
a) Calculate
a k for the sequence an = 2 +
k 60
k 0
16
80
a k and
b) Calculate
n
.
10
a k for the sequence (an) with a0 = 3 and an + 1 = an +
k 40
k 0
1
2
10
6
a k = 22 and
c) Calculate the initial value a0 and the common difference d for the arithmetic sequence (an) with
k 0
a k = 7.
k 0
90
a k = 31 and initial value a0 = 1.
d) Calculate the common difference d for the arithmetic sequence (an) with
k 10
Exercise 12: Geometric series
20
100
a k for the sequence an = 100∙0,9n.
a k and
a) Calculate
k 70
k 0
10
50
a k for the sequence (an) with a0 = 3 and an + 1 = 1,2∙an.
a k and
b) Calculate
k 0
k 40
7
3
a k = 641 and
c) Calculate initial value a0 and common ratio q for the geometric sequence (an) with
k 0
a k = 625.
k 0
4
a k = 336,16 and limit
d) Calculate initial value a0 and common ratio q for the geometric sequence (an) with
k 0
a k = 500
k 0
2
Exercise 13: Limit of a series
n
a k on monotonicity and boundedness. Then draw a conclusion about its convergence:
Examine the series
k 1
n
1
1
1
1
1
= 2 + 2 + 2 + ... + 2
2
1
2
3
n
k 1 k
n
1
1
1
1
1
b)
= 2 + 2 + 2 + ... +
2
1)
1
3
5
(2n 1)2
k 0 (2k
a)
n
1
1
1
1
1
= 1 + 2 + 3 + ... + n
k
3
3
3
3
k 1 3
n
1
1
1
1
1
d)
=
+
+
+ ... +
k
n 3
n 1
n 2
2n
k 1 n
c)
Exercise 14: Mathematical induction
Prove with mathematical induction
a) 2 + 4 + 6 + ... + 2n = n(n + 1) for n ≥ 1
1
b) 12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1) for n ≥ 1
6
1
c) 6 + 24 + 60 + 120 + … + n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)
4
n 1
1 x
d) x0 + x1 + x2 + x3 + .... + xn =
for n ≥ 0.
1 x
1  (n  1)x n  nx n 1
e) 1 + 2x + 3x2 + 4x3 + … + nxn − 1 =
for n ≥ 1
(1  x)2
1
n(n + 1)(n + 2).
3
2n  1
an
3
g) The sequence an mit a1 =
and an + 1 = an −
has the general formula an =
.
(3n  4)(2n  1)
4
3n  1
h) 8n − 1 is divisible by 7 for n ≥ 1
i) n3 − n is divisible by 6 for n ≥ 2
f)
The sequence an with a1 = 2 and an + 1 = an + (n + 1)(n + 2) has the general formula an =
j)
(1 + x)n > 1 + nx for n ≥ 2, x > −1 and x ≠ 0 (Bernoulli’s inequality)
Exercise 15: Monotonicity and Boundedness
a) Show by mathematical induction that the sequence (an)with a0 = 3 and an + 1 =
1
an
2
3
an
is positive for all n ∈ ℕ
1
3
x
> 3 to x and show that the sequence from a) has the lower bound 3 .
2
x
c) Show that the sequence from a) is monotonically decreasing.
d) Find the limit a = lim an of the sequence from a).
b) Solve the inequality
n
Hint: For n → ∞ holds an = an + 1 = lim an = a. Thus in the recursive formula you can plug in an = an + 1 = a and the solve to
n
a.
3
5.7. Solutions to the exercises on sequences and series
Exercise 1: Linear and exponential growth
All lengths are given in dm, all areas in dm2:
a) U0 = 4, U1 = 6, U2 = 8, U3 = 10 and U4 = 12
b) Un+1 − Un = 2 (linear growth
c) Un+1 = Un + 2 (recursive formula)
d) Un = 4 + 2n (general formula)
1
1
1
1
1
1
e) A0 = 1, A1 = 1 + ≈ 1,33, A2 = 1 + +
≈ 1,44, A3 = 1 + +
+
≈ 1,48
9
9
27
3
3
3
1
1
1
1
and A4 = 1 + +
+
+
≈ 1,49
9
27
3
81
f)
An+1 − An = 3n+1∙
1
3
g) An+1 = An +
h) An = 1 +
i)
j)
1
3
1
+
2
1
3n
1
=
1
3
n 1
(exponential growth)
n 1
(recursive formula)
1
3
2
+ ... +
1
3
n
=
1 1/ 3n 1
3
1
=
1 1/ 3
2
1
3
n 1
=
n
3
1 1
−
2
2 3
.(general formula)
U100 = 204 and A100 ≈ 1,5 ⇒ The circumference grows beyond all bounds but the area is bounded!
lim An = lim
n
n
3
1
2
1
3
n 1
=
3
2
Exercise 2: Calculating terms with recursive and general formulae
a) a0 = 100; a1 = 50; a2 = 25; a3 = 12,5; a4 = 6,25 with lim an = 0
n
b) a0 = 50; a1 = 75; a2 = 87,5; a3 = 93,75; a4 = 96,875 with lim an = 100
n
1
1
1
1
; a2 = ; a3 = ; a4 = with lim an = 0
n
3
5
2
4
a0 = 2; a1 = 6; a2 = 12; a3 = 20; a4 = 30 and no limit
a0 = 3; a1 = 3,5; a2 = 4; a3 = 4,5; a4 = 5 and no limit
a0 = 3; a1 = 4,5; a2 = 6,75; a3 = 10,125; a4 = 15,1875 and no limit
a0 = 3; a1 = 4; a2 = 4,5; a3 = 4,75; a4 = 4,875 with lim an = 5
c) a0 = 1; a1 =
d)
e)
f)
g)
n
h) a0 = 3; a1 = 3,3; a2 ≈ 3,74; a3 ≈ 3,98; a4 = 4,18 with lim an = 5
n
Exercise 3: Finding general and recursive formulae from given terms of a sequence
n
1
a) an+1 = an + 2 with a0 = 1 ⇒an = 1 + 2n
e) an+1 = an +
with a0 = 0 ⇒an =
(n 1)(n 2)
n 1
2
2
an with a0 = 1⇒an =
3
3
b) an+1 = 2an with a0 = 3 ⇒an = 3∙2n
f) an+1 =
c) an+1 = 3an with a0 = 2 ⇒an = 2∙3n
g) an+1 = an+
d) an+1 = an + 2n + 3 with a0 = 2 ⇒ an = n2 + 1
h) an+1 =
(2n
n
4n 1
6
with a0=−1 ⇒an=
2n 1
3)(2n 1)
n2
1  n 1

 an for n ≥ 1 with a1 = 1 and a0 = 0 ⇒an = n
3 n 
3
2
Exercise 4: Converting general formulae into general formulae
a) an+1 = an + 3 with a0 = 2
b) an+1 = an + 2n − 1 with a0 = 0
1
an with a0 = 1
3
1
d) an+1 = an +
with a0 = 0
(n 1)(n 2)
c) an+1 =
Exercise 5: Converting recursive formulae into general formulae
a) an = −3n + 2
b) an = 20∙0,8n
c) an = n(n + 1)
d) an = n2
4
Exercise 6: Monotonicity of a sequence
a
n(n 1)
n2 n
a) monotonically increasing, since n 1 =
= 2
> 1 for all n ∈ ℕ.
(n 2)(n 1)
an
n
n 2
b) monotonically increasing, since
c)
an
1
=
an
(n 1) 2
n
2
(n 1)
=
n2
n
2
n
n
n
> 1 for all n ∈ ℕ.
monotonically increasing for n ≥ 2,
since an + 1 − an = (n + 1)3 − 3(n + 1)2 − n3 + 3n2 = 3n2 − 3n − 2 = 3(n2 − n −
2
) > 0 for n ≥ 2
3
d) monotonically decreasing for n ≥ 2,
since an + 1 − an = (n + 1)2∙2− (n + 1) − n2∙2− n = 2− (n+1)(n2 +2n + 1 − 2n2) = 2− (n+1)(− n2 +2n + 1) < 0 for n ≥ 2
Exercise 7: Boundedness of a sequence
a)
Upper bound So = 1, since an ≤ 1 ⇔
n
n 1
≤ 1 for all n ∈ ℕ and lower bound Su = 0, since an ≥ 0 ⇔
b) No upper bound S, since there is no S ∈ ℝ with an ≤ S ⇔
Lower bound Su = 0, since an ≥ 0 ⇔
c)
n2
n2
n
n 1
≥ 0 for n ∈ ℕ
n ≤ S ⇔ n2 + n ≤ S2 for all n ∈ ℕ.
n ≥ 0 for all n ∈ ℕ.
Upper bound So = 0, since an ≤ 0 ⇔ n2 − n3 ≤ 0 ⇔ n2(1 − n) ≤ 0 for all n ∈ ℕ.
No lower bound S, since there is no S ∈ ℝ with an ≥ S ⇔ n2 − n3 ≥ S for all n ∈ ℕ
d) Upper bound So = 1, since an ≤ 1 ⇔ n2∙3− n ≤ 1 ⇔ n2 ≤ 3n for all n ∈ ℕ .
Lower bound Su = 0, since n2∙3− n ≥ 0 for all n ∈ ℕ
Exercise 8: Limit of a sequence
We have to show that for any ε > 0 there is a ne so that the condition ∣aε − a∣ < ε holds for all n > nε.
n 2
1
1
1
a) lim an = 1, since ∣ a − an∣ ≤ ε ⇔ ∣1 −
∣≤ε⇔
≤ ε ⇔ − 1 ≤ n holds for all n ≥ nε = − 1.
n
n 1
n 1
b)
⇔1+
c)
d)
1
1
∣≤ε⇔ 1
≤1− ε
n
n
1
1
≤ n⇔
≤ n holds for all n ≥ nε =
.
(2 )
(2 )
lim an = 1, since ∣ a − an∣ ≤ ε ⇔ ∣1 −
n
1
1
≤ (1 − ε)2 ⇔
n
1 (1
)2
1
n
n2
n ∣≤ε⇔1− 1
1
1
1
sin(n)∣ ≤ ε ⇔
≤ n holds for all n ≥ nε = .
n
n
2n
1
lim an = 0, since ∣a − an∣ ≤ ε ⇔ ∣n3∙2− n ∣ ≤ ε ⇔ ≤ 3 holds for all n ≥ 10
n
n
lim an = 0, since ∣a − an∣ ≤ ε ⇔ ∣
Exercise 9: Convergence of a sequence
1 4n
3
3
3
a) Boundedness: an =
= −2 +
⇒ −2 < an < 1, since an + 2 =
and 0 <
<3
1 2n
1 2n
1 2n
1 2n
3
3
3
3
Monotonicity: an + 1 – an =
−
=
−
= < 0 ⇒ (an) decreases monotonically (to lim a n = −2)
n
1 2(n 1) 1 2n
2n 3 2n 1
n 1
1
1
1
1
1
=
−
⇒ 0 < an < , since 0 <
< .
2n
2n
2n
2
2
2
1
1
1
Monotonicity: an + 1 – an = −
+
> 0 ⇒ (an) increases monotonically (to lim a n = )
n
2n
2
2(n 1)
b) Boundedness: an =
5
n
c) Boundedness: an =
n
2
2n
3
=
3n
3 1
3n
2n
− n
n
2
3
22n
=
1
1
=
n
2
3
n
2n
2n
n
2
3
3n 1
1
1
3n
3n
1
1
1
=
n
2
3
n
2
3
(2n
⇒ −2 < an < 0, since 0 <
6n 32n 1 (22n 1 6n
(2n 1 3n 1 )(2n 3n )
3n 1 )(2n
(2n
32n 1 )
=
1
n
2
3
1
1
n
2
3
1
3n ) (2n 1 3n 1 )(2n
3n 1 )(2n 3n )
(2
n 1
2 32n 1
3n 1 )(2n
3n )
<1+
2
3
n
<2
3n )
>0
⇒ (an) increases monotonically (to lim a n = −1)
n
n n 10
1
d) Boundedness: an =
=
n2
n
10
⇒ 0 < an < 1 + 10 = 11
n2
1
10
−
2
(n 1)
n
1
n 1
1
10
=
n2
1
n 1
n
10
(n 1) 2
10
< 0, since both brackets < 0
n2
⇒ (an) decreases monotonically (to lim a n = 0)
n
Exercise 10: Series and sigma notation
n 1
n
ak
Series
Generating sequence an
Integral
Corresponding function f(x)
k 0
1
1
1
+
+
+…
2
4
3
1
1
1
1+
+
+
+…
3
9
27
1
1
1
1+
+
+
+…
4
9 16
1
n
1
3n
1
x
1
3x
1+
1
n2
1
n(n 1)
f (x)dx
1
1
1
1
1
+
+
+
+…
2
20
6 12
ln(n + 1)
3∙ln(3)∙ 1
1
3n
1
n 1
n 1
ln(2) + ln
n 2
1
x2
1
1
1
=
−
x
x 1
x(x 1)
1−
Exercise 11: Arithmetic series
20
100
a k = 63 and
a)
100
k 60
k 0
16
10
c)
k 0
80
39
ak −
ak =
k 40
k 0
a k = 707 – 297 = 410.
k 0
80
a k = 119 and
b)
59
ak −
ak =
k 0
a k = 1863 – 510 = 1353.
k 0
a k = 11a0 + 55d = 22 ⇔ a0 + 5d = 2 and
k 0
a k = 7a0 + 21d = 7 ⇔ a0 + 3d = 1 result in a0 = −
k 0
90
9
a k = 91∙1 + 4095d =
d)
6
k 0
1
1
and d =
2
2
a k + 31 = 10∙1 + 45d + 31 ⇔ 91 + 4095d = 41 + 45d ⇔ 50 = 4050 d result in d =
k 0
1
81
Exercise 12: Geometric series
20
100
a k = 1000∙(1 – 0,921) ≈ 890,581 and
a)
k 70
k 0
10
50
a k = 15∙(1,211 – 1) ≈ 96,45 and
b)
7
a k = a0∙
k 0
a k = 15∙(1,251 – 1) ≈ 163 792,89
k 40
k 0
c)
a k = 1000∙(1 – 0,9101) ≈ 999,976.
q8 1
= 641 and
q 1
3
a k = a0∙
k 0
q4 1
q8
= 625 result in∙ 4
q 1
q
1
(q 4 1)(q 4
641
=
⇔
q4 1
1 625
1)
= 1,0256
⇔ q4 = 0,256 ⇒ common ratio q = 0,4 and initial value a0 ≈ 384,85
4
336,16
q5 1
1
a k = a0∙
a k = a0 ∙
d)
= 336,16 and limit
= 500 result in q5 – 1 = −
⇔ q5 = 0,32768 ⇔ common
500
1 q
q 1
k 0
k 0
ratio q = 0,8 and initial value a0 = 100.
6
Exercise 13: Limit of a series
n
1
1
a) sn =
is monotonically increasing, since sn+1 − sn =
> 0 for all n ∈ ℕ. (sn) is bounded above, since
2
k
(n 1) 2
k 1
sn =
1
1
+
2
1
2
2
+
1
3
2
n
1
+ ... +
n
<1+
2
1
1
x
1
x
dx = 1 +
2
n
=2−
1
1
< 2 for all n ∈ ℕ. Therefore (sn) converges to a
n
2
1
=
≈ 1,64
lim sn ≤ 2. L. Euler showed in 1736, that lim sn =
2
n
n
6
k 1 k
n
1
1
b) sn =
2
(2k
1)
(2n 3) 2
k 0
sn =
1
1
+
2
1
3
2
+
1
5
2
(2n 1)
Therefore (sn) converges to lim sn ≤
n
n
c)
sn =
k
sn = 1 +
n
1
+ ... +
2
1
<1+
0
(2x 1)
3
. L. Euler showed, that lim sn =
n
2
1
1
is monotonically increasing, since sn+1 − sn = n
k
3
0 3
1
1
3
+
1
3
2
+ ... +
1
3
n
n
n
1
dx =
3x
1
<
1
2(2x 1)
dx = 1 +
2
e
ln(3) x
1
0
2
1
k 0
(2k
1
3
3
−
<
for all n ∈ ℕ.
4n 2
2
2
=
1)
2
=
8
≈ 1,23
1
e
ln(3)
dx =
n
n
ln(3) x
=
1
3
1
3
−
<
for n ∈ ℕ.
n
ln(3)
ln(3)
ln(3) 3
3
Therefore (sn) converges to lim sn ≤
. The exact value of this limit can be obtained by using the summation rule for
n
ln(3)
1
= lim
k
n
0 3
n
1
= lim 1
k
n
0 3
1
3
1
n 1
1
3
= .
1
2
k
k
1
3
n
1
1
1
1
1
d) sn =
+
−
=
2n
1
2n
2
n
1
n
k
2n
1
k 1
geometric series:
(sn) is bounded above, since sn =
1
n 1
+
1
n
2
1
1
3
1
+
n
3
=
1
<
2n
+ ... +
n
0
1
n
x
dx = ln(n
1
2n
2
> 0 for n ∈ ℕ.
n
x) 0 =ln(2n) – ln(n) = ln(2)
for all n ≥ 1 and s0 = 1. Therefore (sn) converges to lim sn ≤ ln(2).
n
Exercise 14: Mathematical induction
a) Base case n = 1: 2 = ∙1∙(1 + 1)
Inductive step n ⇒ n + 1:
Inductive hypothesis for some n: 2 + 4 + 6 +... + 2n = n(n + 1)
We have to show the statement for n + 1: 2 + 4 + 6 +... + 2n + 2(n + 2 = (n + 1)(n + 2)
Plug the hypothesis into the left side:
2 + 4 + 6 +... + 2n + 2(n + 2) = n(n + 1) + 2n + 2 = n 2 + 3n + 2
Right side: (n + 1)(n + 2) = n2 + 3n + 2 qed
1
b) Base case n = 1: 12 = 1(1 + 1)(2 + 1) = 1
6
Inductive hypothesis for some n: 12 + 22 + 32 + ... + n2 =
1
n(n + 1)(2n + 1)
6
We have to show the statement for n + 1: 12 + 22 + 32 + ... + n2 + (n + 1)2 =
1
12 + 22 + 32 + ... + n2 + (n + 1)2 = n(n + 1)(2n + 1) + (n + 1)2 =
6
1
1
Right side: (n + 1)(n + 2)(2n + 3) = (2n3 + 9n2 + 13n + 6) =
6
6
1 3
n +
3
1 3
n +
3
1
(n + 1)(n + 2)(2n + 3)
6
3 2 13
n +
n+1
6
2
3 2 13
n +
n + 1 = left side, qed
6
2
7
c)
Base case n = 1: 6 =
1
1‧2‧3‧4 = 6
4
Inductive hypothesis for some n: 6 + 24 + 60 + 120 + … + n(n + 1)(n + 2) =
1
n(n + 1)(n + 2)(n + 3)
4
We have to show the statement for n + 1:
6 + 24 + 60 + 120 + … + n(n + 1)(n + 2) + (n + 1)(n + 2)(n + 3) =
1
(n + 1)(n + 2)(n + 3)(n + 4)
4
6 + 24 + 60 + 120 + … + n(n + 1)(n + 2) + (n + 1)(n + 2)(n + 3)
1
n(n + 1)(n + 2)(n + 3) + (n + 1)(n + 2)(n + 3)
4
1
= (n + 1)(n + 2)(n + 3)(n + 4) = right side, qed.
4
=
1  x1
=1
1 x
d) Base case n = 0: 1 =
Inductive hypothesis for some n: x0 + x1 + x2 + x3 + .... + xn =
1 xn 1
1 x
We have to show the statement for n + 1: x0 + x1 + x2 + x3 + .... + xn + xn+1 =
1 xn 2
1 x
x0 + x1 + x2 + x3 + .... + xn + xn+1 =
e)
1 xn
1 xn 1
+ xn + 1 =
1 x
1
xn 1
1 x
xn
2
=
1 xn 2
= right side, qed
1 x
1  2x  x 2
=1
(1  x) 2
Inductive hypothesis for some n: 1 + 2x + 3x2 + 4x3 + … + nxn − 1 =
1  (n  1)x n  nx n 1
(1  x)2
We have to show the statement for n + 1: 1 + 2x + 3x2 + 4x3 + … + nxn − 1 + (n + 1)xn =
1  (n  2)x n 1  (n  1)x n  2
(1  x)2
1  (n  1)x n  nx n 1
1  (n  1)x n  nx n 1  (1  x)2 (n  1)x n
+ (n + 1)xn
=
2
(1  x)
(1  x)2
f)
=
1  (n  1)x n  nx n 1  (n  1)x n  2(n  1)x n 1  (n  1)xn  2
(1  x)2
=
1  (n  2)x n 1  (n  1)x n  2
= right side, qed
(1  x)2
1
‧1‧2‧3 = 2
3
1
n(n + 1)(n + 2)
3
1
We have to show the statement for n + 1: an + 1 = (n + 1)(n + 2)(n + 3)
3
1
1
2
1
11
:an + 1 = an + (n+ 1)(n + 2) = n(n + 1)(n + 2) + (n + 1)(n + 2) = n3 + n2 + n + n2 + 3n + 2 = n3 + 2n2 +
n + 2.
3
3
3
3
3
1
1
1
11
Right side: (n + 1)(n + 2)(n + 3) = (n2 + 3n + 2)(n + 3) = n3 + 2n2 +
n + 2 = left side, qed.
3
3
3
3
Inductive hypothesis for some n: an =
8
g) Base case for n = 1:
2 1
3
3
=
=
3 1
4
4
Inductive hypothesis for some n: an =
2n  1
3n  1
We have to show the statement for n + 1: an + 1 =
2n  3
3n  4

an
2n  1 
1
2n  1 6n 2  11n  3
2n  1 (3n  1)(2n  3)
a n + 1 = an −
=
=
=
 1 


 =
(3n  4)(2n  1)
3n  1 (3n  4)(2n  1)
3n  1  (3n  4)(2n  1) 
3n  1 (3n  4)(2n  1)
2n  3
= right side, qed.
3n  4
h) Base case n = 1: 81 − 1 = 7 obviously is divisible by 7
i)
Inductive hypothesis for some n: 8n −1 is divisible by 7
We have to show the statement for n + 1: 8n + 1 − 1 = 8∙8n − 1 = 7∙8n + 8n – 1 is divisible by 7
Obviously the first summand 7∙8n 1 is divisible by 7. According to the inductive hypothesis the second summand 8n − 1
too is divisible by 7 and therefore the complete sum is divisible by 7. qed
Base case n = 1: 23 − 2 = 6 1 is divisible by 6
Inductive hypothesis for some n: n3 − n is divisible by 6
We have to show the statement for n + 1:
(n + 1)3 − (n + 1) = n3 + 3n2 + 2n = (n3 − n) + (3n2 + 3n) = (n3 − n) + 6∙
1
n(n + 1) is divisible by 6
2
According to the inductive hypothesis the left summand n3 − n is divisible by 6. But since either n or n + 1 is even,
1
2
1
n(n + 1) too is divisible by 6 and so is the complete sum. qed
2
Base case n = 2: (1 + x)2 = 1 + 2x + x2 > 1 + 2x, since x2 > 0∙
n(n + 1) is an integer. Therefore the right summand 6∙
j)
Inductive hypothesis for some n: (1 + x)n > 1 + nx
We have to show the statement for n + 1: (1 + x)n + 1 > 1 + (n + 1)x
(1 + x)n + 1 (1 + x)n(1 + x) > (1 + nx)(1 + x) , since we assume 1 + x > 0
= 1 + (n + 1)x + x2
> 1 + (n + 1)x, since x2 > 0
= right side, qed
Aufgabe 15: Mathematical induction, Monotonicity and Boundedness
a) Base case n = 0: a0 = 3 > 0∙
Inductive hypothesis for some n: an > 0
We have to show the statement for n + 1: an + 1 > 0
Use the hypothesis: an + 1 =
b)
1
x
2
c) an
+ 1
3
>
x
– an =
3 ⇔x+
1
an
2
1
an
2
3
>2 3
x
3
an
− an =
3
an
> 0, since an > 0, qed.
x 0 because of a)
1 3
2 an
an
x2 + 3 > 2 3 x ⇔ x2 − 2 3 x + 3 > 0 ⇔ (x −
=
1 3 an2
2 an
3 ) > 0 for x ∈ ℝ.
> 0, because in b) we have shown an2 < 3 ⇒ (an) is
monotonically decreasing.
d) For n → ∞ we have an = an + 1 = lim an = a and so a =
n
1
a
2
3
1
1 3
⇔ a=
⇔ a2 = 3
a
2
2 a
a 0 because of a)
a = lim an =
n
3.
9

5.7. Exercises on sequences and series - Poenitz

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