Bài tập chuỗi só Lê Xuân Đại (ĐH Bách Khoa HCM)

a
0
+ a
1
+ a
2
+ . . . + a
n
+ . . . ,
a
i
i = 0, 1, 2, . . . , n, . . .
+∞

n =0
a
n
.
n
0
∈ N.
+∞

n =n
0
a
n
a
0
+ a

1
+ a
2
+ . . . + a
n
+ . . . ,
a
i
i = 0, 1, 2, . . . , n, . . .
+∞

n =0
a
n
.
n
0
∈ N.
+∞

n =n
0
a
n
S
n
=
n

k=0

a
k
= a
0
+ a
1
+ a
2
+ . . . + a
n
+∞

n=0
a
n
.
+∞

n=0
a
n
S

S
n

+∞
n=1
. S
+∞


n=0
a
n
.
S
n
=
n

k=0
a
k
= a
0
+ a
1
+ a
2
+ . . . + a
n
+∞

n=0
a
n
.
+∞

n=0

a
n
S

S
n

+∞
n=1
. S
+∞

n=0
a
n
.
1 +
1
2
+
1
4
+ . . . +
1
2
n
+ . . . =
+∞

n =0

1
2
n
.
S
n
= 1+
1
2
+
1
4
+. . .+
1
2
n
=
1 −
1
2
n+1
1 −
1
2
= 2

1 −
1
2
n +1


lim
n →+∞
S
n
= lim
n →+∞
2

1 −
1
2
n +1

= 2.
1 +
1
2
+
1
4
+ . . . +
1
2
n
+ . . . =
+∞

n =0
1

2
n
.
S
n
= 1+
1
2
+
1
4
+. . .+
1
2
n
=
1 −
1
2
n+1
1 −
1
2
= 2

1 −
1
2
n +1


lim
n →+∞
S
n
= lim
n →+∞
2

1 −
1
2
n +1

= 2.
1 +
1
2
+
1
4
+ . . . +
1
2
n
+ . . . =
+∞

n =0
1
2

n
.
S
n
= 1+
1
2
+
1
4
+. . .+
1
2
n
=
1 −
1
2
n+1
1 −
1
2
= 2

1 −
1
2
n +1

lim

n →+∞
S
n
= lim
n →+∞
2

1 −
1
2
n +1

= 2.
+∞

n =0
a
n

S
n

+∞
n =1
n → +∞,
+∞

n =0
q
n

, q ∈ R.
+∞

n =0
a
n

S
n

+∞
n =1
n → +∞,
+∞

n =0
q
n
, q ∈ R.
S
n
= 1 + q + q
2
+ . . . + q
n
=



1 − q

n+1
1 − q
, q = 1
n + 1, q = 1
|q| = 1 lim
n→∞
S
n
= lim
n→∞
1 − q
n+1
1 − q
=
lim
n→∞

1
1 − q
−
q
n+1
1 − q

=



1
1 − q

, |q| < 1
∞, |q| > 1
q = 1 lim
n→∞
S
n
= lim
n→∞
n + 1 = ∞
q = −1 lim
k→∞
S
2k+1
= 0, lim
k→∞
S
2k
= 1.
|q| < 1
∞

n=0
q
n
=
1
1 − q
.
S
n

= 1 + q + q
2
+ . . . + q
n
=



1 − q
n+1
1 − q
, q = 1
n + 1, q = 1
|q| = 1 lim
n→∞
S
n
= lim
n→∞
1 − q
n+1
1 − q
=
lim
n→∞

1
1 − q
−
q

n+1
1 − q

=



1
1 − q
, |q| < 1
∞, |q| > 1
q = 1 lim
n→∞
S
n
= lim
n→∞
n + 1 = ∞
q = −1 lim
k→∞
S
2k+1
= 0, lim
k→∞
S
2k
= 1.
|q| < 1
∞


n=0
q
n
=
1
1 − q
.
S
n
= 1 + q + q
2
+ . . . + q
n
=



1 − q
n+1
1 − q
, q = 1
n + 1, q = 1
|q| = 1 lim
n→∞
S
n
= lim
n→∞
1 − q
n+1

1 − q
=
lim
n→∞

1
1 − q
−
q
n+1
1 − q

=



1
1 − q
, |q| < 1
∞, |q| > 1
q = 1 lim
n→∞
S
n
= lim
n→∞
n + 1 = ∞
q = −1 lim
k→∞
S

2k+1
= 0, lim
k→∞
S
2k
= 1.
|q| < 1
∞

n=0
q
n
=
1
1 − q
.
+∞

n=1
1
n(n + 1)
S
n
=
1
1.2
+
1
2.3
+ . . . +

1
n(n + 1)
.
1
n(n + 1)
=
1
n
−
1
n + 1
, n ∈ N.
S
n
=
1
1
−
1
2
+
1
2
−
1
3
+ . . . +
1
n
−

1
n + 1
= 1 −
1
n + 1
.
lim
n→∞
S
n
= lim
n→∞

1 −
1
n + 1

= 1.
+∞

n=1
1
n(n + 1)
S
n
=
1
1.2
+
1

2.3
+ . . . +
1
n(n + 1)
.
1
n(n + 1)
=
1
n
−
1
n + 1
, n ∈ N.
S
n
=
1
1
−
1
2
+
1
2
−
1
3
+ . . . +
1

n
−
1
n + 1
= 1 −
1
n + 1
.
lim
n→∞
S
n
= lim
n→∞

1 −
1
n + 1

= 1.
+∞

n=1
1
n(n + 1)
S
n
=
1
1.2

+
1
2.3
+ . . . +
1
n(n + 1)
.
1
n(n + 1)
=
1
n
−
1
n + 1
, n ∈ N.
S
n
=
1
1
−
1
2
+
1
2
−
1
3

+ . . . +
1
n
−
1
n + 1
= 1 −
1
n + 1
.
lim
n→∞
S
n
= lim
n→∞

1 −
1
n + 1

= 1.
+∞

n=1
1
n(n + 1)
S
n
=

1
1.2
+
1
2.3
+ . . . +
1
n(n + 1)
.
1
n(n + 1)
=
1
n
−
1
n + 1
, n ∈ N.
S
n
=
1
1
−
1
2
+
1
2
−

1
3
+ . . . +
1
n
−
1
n + 1
= 1 −
1
n + 1
.
lim
n→∞
S
n
= lim
n→∞

1 −
1
n + 1

= 1.
+∞

n=1
1
n(n + 1)
S

n
=
1
1.2
+
1
2.3
+ . . . +
1
n(n + 1)
.
1
n(n + 1)
=
1
n
−
1
n + 1
, n ∈ N.
S
n
=
1
1
−
1
2
+
1

2
−
1
3
+ . . . +
1
n
−
1
n + 1
= 1 −
1
n + 1
.
lim
n→∞
S
n
= lim
n→∞

1 −
1
n + 1

= 1.
+∞

n =1
a

n
lim
n →+∞
a
n
= 0.
+∞

n =1
a
n
⇔ lim
n →+∞
S
n
= S.
lim
n →+∞
a
n
= lim
n →+∞
(S
n
− S
n −1
) =
= lim
n →+∞
S

n
− lim
n →+∞
S
n −1
= S − S = 0
lim
n→+∞
a
n
= 0
+∞

n=1
a
n
+∞

n =1
a
n
lim
n →+∞
a
n
= 0.
+∞

n =1
a

n
⇔ lim
n →+∞
S
n
= S.
lim
n →+∞
a
n
= lim
n →+∞
(S
n
− S
n −1
) =
= lim
n →+∞
S
n
− lim
n →+∞
S
n −1
= S − S = 0
lim
n→+∞
a
n

= 0
+∞

n=1
a
n
+∞

n =1
a
n
lim
n →+∞
a
n
= 0.
+∞

n =1
a
n
⇔ lim
n →+∞
S
n
= S.
lim
n →+∞
a
n

= lim
n →+∞
(S
n
− S
n −1
) =
= lim
n →+∞
S
n
− lim
n →+∞
S
n −1
= S − S = 0
lim
n→+∞
a
n
= 0
+∞

n=1
a
n
+∞

n =1
a

n
lim
n →+∞
a
n
= 0.
+∞

n =1
a
n
⇔ lim
n →+∞
S
n
= S.
lim
n →+∞
a
n
= lim
n →+∞
(S
n
− S
n −1
) =
= lim
n →+∞
S

n
− lim
n →+∞
S
n −1
= S − S = 0
lim
n→+∞
a
n
= 0
+∞

n=1
a
n
+∞

n=1
1
√
n
lim
n→+∞
a
n
= lim
n→+∞
1
√

n
= 0.
S
n
= 1 +
1
√
2
+ . . . +
1
√
n
 n.
1
√
n
=
√
n, n ∈ N
lim
n→+∞
S
n
 lim
n→+∞
√
n = +∞ ⇒ lim
n→+∞
S
n

= +∞
+∞

n=1
1
√
n
+∞

n=1
1
√
n
lim
n→+∞
a
n
= lim
n→+∞
1
√
n
= 0.
S
n
= 1 +
1
√
2
+ . . . +

1
√
n
 n.
1
√
n
=
√
n, n ∈ N
lim
n→+∞
S
n
 lim
n→+∞
√
n = +∞ ⇒ lim
n→+∞
S
n
= +∞
+∞

n=1
1
√
n
+∞


n=1
1
√
n
lim
n→+∞
a
n
= lim
n→+∞
1
√
n
= 0.
S
n
= 1 +
1
√
2
+ . . . +
1
√
n
 n.
1
√
n
=
√

n, n ∈ N
lim
n→+∞
S
n
 lim
n→+∞
√
n = +∞ ⇒ lim
n→+∞
S
n
= +∞
+∞

n=1
1
√
n