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a : N→ R
a(n) a n ∈ N an n
(an)n∈N (an)
(an)n∈N M
|an| M n ∈ N
a ∈ R
(an)n∈N a = limn→+∞
an ε > 0 n0
n > n0 |an − a| < ε
a = +∞ (an)n∈N
limn→+∞
an =
+∞ M n0
n > n0 an > M lim
n→+∞an = −∞
M n0
n > n0 an < M
a = limn→+∞
an a ∈ R a = ±∞ (an)n∈N
a n +
∞
(an)n∈N a
a
a (an)n∈N
a
a = ±∞ (an)n∈N
+∞ −∞
R
(an) (bn) limn→+∞
an = a
limn→+∞
bn = b
1◦ limn→+∞
(an ± bn) = a ± b
2◦ limn→+∞
(anbn) = ab
3◦ limn→+∞
an
bn=
a
b bn = 0
n ∈ N b = 0
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R
lim
n→+∞an = a
limn→+∞
bn = b a < b an < bn
n lim
n→+∞an = a a < b an < b
n < >
n ∈ N
1
n2 <
1
n
limn→+∞
1
n2 = lim
n→+∞
1
n = 0
n n0 an bn (an)n∈N (bn)n∈N limn→+∞an
limn→+∞
bn
(an)n∈N
ε > 0 n0 |am−an| < ε
m n n0
(an)n∈N n ∈ N
an an+1 n ∈ N an < an+1
(an)n∈N n ∈ N an an+1
n ∈ N an > an+1
1◦ limn→+∞
1
na = 0
a > 0
2◦ limn→+∞
na = +∞ a > 0
3◦ limn→+∞
q n = 0 |q | < 1
4◦ limn→+∞
q n = +∞ q > 1
5◦ limn→+∞
nk
q n = 0
|q | > 1
k ∈ N
7/17/2019 Brojni redovi
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6◦ limn→+∞
q n
nk
= 0
|q
| 1 k
∈N
7◦ limn→+∞
q n
n! = 0 q ∈ R
8◦ limn→+∞
n
√ a = 1 a > 0
9◦ limn→+∞
n
√ n = 1
10◦ limn→+∞
1 +
1
n
n= e
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