Lista 6, Kierunek: AiR, sem. I, 2008/2009
Granice ciagów liczbowych
1. Oblicz granice nastepujacych ciagów liczbowych:
(n20+2)3 (n2+1)n!+1
n3+2n2+1
a) an = b) bn = , c)cn = ,
3
(n3+1)20 (2n+1)(n+1)!
"n-3n "
" "
3
8n+1+3 n3+1
"
d) dn = , e) an = , f) bn = n2 + 4n + 1 - n2 + 2n,
5
2n+1
n5+1+1
"
" "
4 1+3+...+(2n-1)
g) cn = n + 6 n + 1- n, h) dn = n4 + 16-n, i) an = ,
2+4+...+2n
1 1 1 "
1+ + +...+ n
arctg(3n+1)
2
22 2n
"4 +1
j) bn = , k) , l) cn = ,
1 1 1 3
arctg(2n+1)
1+ + +...+ 8n+1
3
32 3n
" "
3 3
2
m) an = n3 + 3n - n, n) bn = n3 + 2n2 - n, o) cn = (-1)n · ( )n
3
1 1 1
1+ + +...+
2 arctgn
22 2n
p) an = , r) bn = .
1+3+...+(2n-1) arcctgn
2. Oblicz granice nastepujacych ciagów korzystajac z twierdzenia o trzech
ciagach:
"
2n+(-1)n
n 2n sin n 3n+2n
n
a) an = n2n + 1, b) bn = , c)cn = , d) an = ,
3n+1 3n+2 5n+4n
" " "
n+1 n n+2
e) bn = 2n + 3, f) cn = 3 + sin n, g) an = 3n + 4n+1,
"
n
n 1 2 3 4
h) bn = 1 + 5n2 + 3n5, i) cn = + + + ,
n n2 n3 n4
nĄ
sin
1 1 1 2n+100
2
j) an = + + . . . + , k) bn = , l) cn =
n2+1 n2+2 n2+n n 3n+1
2n2+sin n!
m) an = .
4n2-3 cos n2
3. Oblicz granice nastepujacych ciagów korzystajac z definicji liczby e:
1 1 1
a)an = (1 + )5n, b) bn = (1 + )-3n, c) cn = (1 + )3n-2,
n n n
n2 2 3n
d) an = (5n+2 )15n, e) an = ( )n , f) an = ( )n, g) an = (3n+1 )6n,
5n+1 n2+1 3n+1 3n+2
2 2
n n+4
h) an = (n+1)n, h) an = ( )5-2n, i) an = (n -1 )2n -3,
n+3 n2
(-1)n n
3n+2 5n+3
j) an = (1 + )(-1) n k) bn = ( )n · ( )n.
n 5n+2 3n+1