musterlösung übung 2 20/21 PDF

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0

(an ) (an )

ǫ>0 √ |a n  | < 2ǫ 2an − an  < ǫ

n0 ∈ N

n ≥ n0 |an · an+1| < ǫ |an · am | < ǫ

m∈N

(an )

ǫ>0 n∈N

ǫ˜ := ǫ2 > 0 (an )

(a) √ |an − 0| = |an | < ǫ˜ = ǫ

n ≥ n0

n0 ∈ N

0

(an ) (an ) an :=

n ≥ n0

an · an+1 =

1 n

(

1 n2

,

n∈N n∈N

n

.

ǫ>0

n∈N |an · an+1| =

n0 ∈ N

n0 >

1 ǫ

1 1 ≤ < ǫ. n0 n

(an )

(an ) (c)

an := 2 (n ∈ N) (an )

|an · am | < ǫ

(d) m∈N |an · an | < ǫ

ǫ>0

⇔ (a)

2

|an | < ǫ

n0 ∈ N

⇔

|an | <

m=n √ ǫ.

n ≥ n0

(an ) √ an := 4n2 + n + 5 − 2 an := (1 + 2(−1)n )n

an := an :=

(n+2)3 −(n−1)3 (n−1)2 +2n2 +5 1+2+···+n 1+3+···+(2n−1)

(an )

an ≥

√ 4n2 − 2 = 2n − 2 = 2(n − 1)

(n)

n ∈ N.

(an ) (an )

3

n3 + 6n2 + 12n + 8 − (n3 − 3n2 + 3n − 1) 9n2 + 9n + 9 = 2 2 2 n − 2n + 1 + 2n + 5 3n − 2n + 6 9 + 9n + n92 n→∞ 9 = −−−−→ = 3, 3 3 − 2n + n62

an =

(an )

(an ) ∀s > 0 ∃n ∈ N : an > s. s>0

k > 4s

k∈N

n := 2k ∈ N

an = (1 + 2(−1)n )n = (1 + 2(−1)2k )2k = (1 + 2)2k ≥ 1 + 4k > 4k > s,

1 2

(an ) Pn

k=1

k=

n(n+1) 2

n∈N

Pn Pn n(n+1) k k 1 n2 + n 2 = n(n+1) = = Pn k=1 Pn an = Pn k=1 1 2 k=1 k − k=1 2 n2 2 2 −n k=1 (2k − 1) =

1 1 + n1 n→∞ 1 −−−−→ . 2 2 1

1 |bn | = 1 (n ∈ N)

an := 1 (n ∈ N) bn := (−1)n (n ∈ N) (bn ) cn := an · bn = (−1)n (n ∈ N)

(cn )

B∈R (an )

(an ) n∈N |an | ≤ A (cn ) n∈N

(bn ) cn := an · bn (n ∈ N) A∈R n∈N

(an )

|bn | ≤ B |cn | = |an | |bn | ≤ A·B =: C

(bn )

|bn | ≤ B B∈R cn := an · bn (n ∈ N) n0 ∈ N |an | < Bǫ

(cn )

n∈ N

ǫ>0

(an )

n ≥ n0 |cn | = |an | |bn | ≤ (cn )

ǫ ·B =ǫ B

n ≥ n0 ,

0

an := n2 (n ∈ N) cn := an · bn = n (n ∈ N)

bn :=

1 n

(bn )

(n ∈ N)

(an )∞ n=0 a0 := 0, a1 := 1, an := ) an = 23 (1 − (−1) 2n

1 (an−1 + an−2 ) 2

n ∈ N, n ≥ 2.

n

A⊆R an ∈ A

n ∈ N0 (an )

limn→∞ an = sup A

n∈N

(an ) (bn ) cn := max{an , bn } (n ∈ N)

a

b

max{a, b}

(an )∞ n=0 a0 := 0, a1 := 1, an := an =

2 3



1−

(−1)n 2n



1 (an−1 + an−2 ) 2 n∈N

n ∈ N, n ≥ 2. ∞ (an )n=0

2 3

n=0

a0 = 0 =

2 3

n+1

n



1−

(−1)0 20

n∈N



n=1

a1 = 1 =   n 2 an = 3 1 − (−1) n 2

2 3



1−

(−1)1 21

an−1 =

 2 3



1−

(−1)n−1 2n−1



     2 (−1)n (−1)n−1 1 1 2 + 1− n an+1 = (an + an−1 ) = 1− 2 2n−1 2 3 2 3    n+1 n−1 n 2(−1) 4(−1)n−1 4(−1) 1 1 2(−1) = 2 + − + 1 − = 1− 2n+1 3 2n+1 2n+1 2n+1 3   n+1  n+1 n+1  (−1) (−1) 2(−1) 2 2 = 1− 1+ − . = 3 3 2n+1 2n+1 2n+1 n∈N     n    an − 2  = − 2 · (−1)  = 2  3  3 2n  3 2 3

∞ (an )n=0

A⊆R

(an )

   1 2 1  ·  n  ≤ · → 0 2 3 n

an ∈ A

n∈N

1 n

limn→∞ an = sup A

sup A ∈ R n∈N an ∈ A

A ǫn :=

(n → ∞),

>0

n ∈ N

an > sup A − ǫn . A

an ≤ sup A

n ∈ N

|an − sup A| ≤ ǫn n∈N

ǫ>0

kǫ ∈ N |an − sup A| ≤ ǫn =

(an )

n ≥ kǫ (an )

1 1

1 ǫ...