musterlösung übung 1 20/21 PDF

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x≤5+

x∈R

√ x+7

A := {(−1)n −

3 n

|x + 5| ≤ 2(4 − x)

: n ∈ N}

B := {−x −

x≤5+

x ∈ [−7, 9] x∈R

x+7≥0 x ≥ −7 x≤5+

√ x+7

⇔

⇔

x ≥ −7

x−5≤

√ x + 7.

x0 sup A ≥ 1 ǫ>0

A sup A = 1

n := 2k (−1)n −

inf B

B r < −2

3 2ǫ

3 3 3 =1− = (−1)2k − > 1 − ǫ. 2k n 2k

max B = sup B = −2 min B min B

k>

k∈N

inf B − 1r ⊆ (0, 12 ) ⊆ (0, 2]

B ∋ −x −

x = − 1r

1 1 ≤ − = r, x x

B −x − x1 ≤ −x ≤ 0 B x ∈ (0, 2] x + 1x ≥ 1 + 11 = 2 x ∈ (0, 2] x ∈ (0, 2] sup B = max B = −2 x+ 0 xy(y + ) ⇔ x + > y + . x y x y

x+ x1 ≥ 1+ 11 = 2 x ∈ (0, 1] < y+ 1≤x 0 x + 1x ≥ 1 + 11 = 2 −x − x1 ≤ −2 sup B = max B = −2

1 x

⇔

x ∈ [1, 2] x ∈ (0, 2]

x+

−x− x1 ≤ −2

x ∈ (0, 1]

1 1 0 a0 > sup A − ε2

a0 ∈ A

a0 + b0 > sup A + sup B − 2 ·

ε b0 ∈ B

b0 > sup B −

ε 2

ε = sup A + sup B − ε = c, 2

sup A + sup B

A+B

A A+B sup(A + B) = sup A + sup B

B

a0 ∈ A

b∈B b = a0 + b − a0 ≤ sup(A + B) − a0 ,

B

∅= 6 A⊆R

sup B =

B 6= ∅ a0 6= 0

a0 ≥ inf A > 0 b ∈ B B 1 inf A 1 S

> a ≥ inf A sup B = inf1 A

B := {b ∈ R :

inf A > 0 1 inf A

b 6= 0

1 b

A= 6 ∅

∈ A

1 inf A

1 a

∈B

1 S

> inf A S < a1 = b

∈ A}

B

a0 ∈ A 1 ∈B a0 1 b

b ≤ B S ∈ (0, inf1 A )

≥ inf A

B

B b :=

1 b

S

1 inf A

B

S a∈A

n∈N n X j=1

1 j 3 = n2 (n + 1)2 . 4

n∈N n X

k=1

n∈N n∈N

3n > 2n3

n≥6 n3 +5n 6

Pn

n∈N j=1

n+1

n+1

X

j=1

P1

n=1 n

j3 =

n X

=

k=1

n+1 n+1 X k=1

k2k =

n X

1 4

· 12 · (1 + 1)2 Pn

j=1

j 3 = 41 n2 (n + 1)2

n+1

k2k = 2 = 0 · 22 + 2P n k n+1 +2 n∈N k=1 k2 = (n − 1)2

(n − 1 + n + 1) + 2 = 2n+2 n + 2 = ((n + 1) − 1)2(n+1)+1 + 2.

n∈N n=6

k2k = (n − 1)2n+1 + 2

k2k + (n + 1)2n+1 = (n − 1)2n+1 + 2 + (n + 1)2n+1

k=1 n+1

=2

Pn

k=1

P1

n=1

n

j3 = 1 = n∈N

1 4 1 1 (n + 6n3 + 13n2 + 12n + 4) = (n2 + 2n + 1)(n2 + 4n + 4) = (n + 1)2 ((n + 1) + 1)2 . 4 4 4

n∈N

n

j 3 = 14 n2 (n + 1)2

1 1 2 n (n + 1)2 + (n + 1)3 = (n4 + 2n3 + n2 ) + n3 + 3n2 + 3n + 1 4 4

j 3 + (n + 1)3 =

j=1

j=1

k2k = (n − 1)2n+1 + 2.

n≥6

3n > 2n3

36 = 729 > 432 = 2 · 63 n∈N n≥6

3n > 2n3

(6≤n)

2(n + 1)3 = 2n3 + 6n2 + 6n + 2 ≤ 2n3 + n3 + n2 + n = 3n3 + n2 + n ≤ 3n3 + 2n3 ≤ 6n3 = 3 · 2n3 < 3 · 3n = 3n+1, n ≤ n2

n∈N

n∈N n=1 n n+1 3n(n + 1) + 6

13 +5·1 6

n3 +5n 6

=1∈N n∈N

n

n3 +5n 6

n+1

∈N

3n(n + 1)

6

n3 + 3n2 + 3n + 1 + 5n + 5 n3 + 5n 3n(n + 1) + 6 (n + 1)3 + 5(n + 1) + = = ∈ N. 6 6 6 } 6 | {z ∈ N...