Lista 6,

Kierunek: AiR, sem. I, 2008/2009

Granice ci

,

ag´

ow liczbowych

1.

Oblicz granice nast

,

epuj

,

acych ci

,

ag´

ow liczbowych:

a) a

n

=

n

3

+2n

2

+1

n−3n

3

b) b

n

=

(n

20

+2)

3

(n

3

+1)

20

,

c)c

n

=

(n

2

+1)n!+1

(2n+1)(n+1)!

,

d) d

n

=

3

√

8

n+1

+3

2

n

+1

,

e) a

n

=

√

n

3

+1

5

√

n

5

+1+1

,

f) b

n

=

√

n

2

+ 4n + 1 −

√

n

2

+ 2n,

g) c

n

=

p

n + 6

√

n + 1−

√

n,

h) d

n

=

4

√

n

4

+ 16−n,

i) a

n

=

1+3+...+(2n−1)

2+4+...+2n

,

j) b

n

=

1+

1
2

+

1

22

+...+

1

2n

1+

1
3

+

1

32

+...+

1

3n

,

k)

√

4

n

+1

3

√

8

n

+1

,

l) c

n

=

arctg(3n+1)
arctg(2n+1)

,

m) a

n

=

3

√

n

3

+ 3n − n,

n) b

n

=

3

√

n

3

+ 2n

2

− n,

o) c

n

= (−1)

n

· (

2
3

)

n

p) a

n

=

1+

1
2

+

1

22

+...+

1

2n

1+3+...+(2n−1)

,

r) b

n

=

arctgn

arcctgn

.

2. Oblicz granice nast

,

epuj

,

acych ci

,

ag´

ow korzystaj

,

ac z twierdzenia o trzech

ciag

,

ach:

a) a

n

=

n

√

n2

n

+ 1,

b) b

n

=

2

n

sin n

3

n

+1

,

c)c

n

=

2n+(−1)

n

3n+2

,

d) a

n

=

n

q

3

n

+2

n

5

n

+4

n

,

e) b

n

=

n+1

√

2n + 3,

f) c

n

=

n

√

3 + sin n,

g) a

n

=

n+2

√

3

n

+ 4

n+1

,

h) b

n

=

n

√

1 + 5n

2

+ 3n

5

,

i) c

n

=

n

q

1

n

+

2

n

2

+

3

n

3

+

4

n

4

,

j) a

n

=

1

n

2

+1

+

1

n

2

+2

+ . . . +

1

n

2

+n

, k) b

n

=

sin

nπ

2

n

,

l) c

n

=

2

n

+100

3

n

+1

m) a

n

=

2n

2

+sin n!

4n

2

−3 cos n

2

.

3. Oblicz granice nast

,

epuj

,

acych ci

,

ag´

ow korzystaj

,

ac z definicji liczby e:

a)a

n

= (1 +

1

n

)

5n

,

b) b

n

= (1 +

1

n

)

−3n

,

c) c

n

= (1 +

1

n

)

3n−2

,

d) a

n

= (

5n+2
5n+1

)

15n

, e) a

n

= (

n

2

n

2

+1

)

n

2

,

f) a

n

= (

3n

3n+1

)

n

,

g) a

n

= (

3n+1
3n+2

)

6n

,

h) a

n

= (

n

n+1

)

n

,

h) a

n

= (

n+4
n+3

)

5−2n

,

i) a

n

= (

n

2

−1

n

2

)

2n

2

−3

,

j) a

n

= (1 +

(−1)

n

n

)

(−1)

n

n

k) b

n

= (

3n+2
5n+2

)

n

· (

5n+3
3n+1

)

n

.