BÀI TẬP CHUỖI LŨY THỪA
Bài tập
n
1. Tìm bán kính hội tụ của các chuỗi sau:
n
2
( + -
n
+ 1
) 1
n
(
)
3
2
( (
=
) 1 ) n 2 !!
n
=
0
(cid:0) (cid:0) - n n 2 - b x (cid:0) ) 2 a x (cid:0) )
n
2
- n n
n
2
n
1
n
(cid:0) (cid:0) -
(
) 1
- - c (cid:0) d ) (cid:0) )
n
n
= 1
= 1
n 1 � � x 1 � � n � �
+ n
3
1
2
n
n
- (cid:0) (cid:0)
(
) 1
n
+ 1
=
f x (cid:0) ) + e x (cid:0) )
n
2
=
n
2
n + x n 3 . 22 n -� � n 1 � �+� � n 2 n 8 3ln
n
n
2
( + -
(cid:0)
Hướng dẫn ) 1
n
n 2
3
2
=
x a (cid:0) )
n
2
n
2/3
1/2
- n n
n
- n n = = R (cid:0) (cid:0) (cid:0) (cid:0) lim n lim n | | 1 a n
n 1 � � + -� � 2 � �
n
2
/3
/ 1 2
2 1 n
n
/ 1
- n = =
n
(cid:0) (cid:0) lim n 1 2
n �� � 1 �� � 2 � � � � � + -� 2 1 � �
n
+ 1
(cid:0) - n
(
)
( (
=
- b x (cid:0) ) 2
) 1 ) !!
n
0
n 2
(
) 2 !!
(
+ 1
+ - n 2 = = R . (cid:0) (cid:0) (cid:0) (cid:0) lim n lim n n n 1 ) n 2 !! a n a n
)
( . 2
- n 1 + = +(cid:0) n 2 (cid:0) (cid:0) = lim n n
n
1
n
(cid:0) -
(
) 1
n
= 1
- - c (cid:0) )
2 � � x 1 � � n � �
+ 1
- n 2 = = = R . 1 (cid:0) (cid:0) (cid:0) (cid:0) lim n lim n - + n n n 1 1 a n a n
n
2
n
2
(cid:0) (cid:0)
= (cid:0) d (cid:0) ) a x n
n
n
= 1
= 1
n x n 3 .
n
= = n R 3= (cid:0) (cid:0) (cid:0) (cid:0) lim 3n n lim n
1 a n
2
n
(cid:0)
(
) 1
n
+ 1
+ e x (cid:0) )
=
n
2
n
n
+ 1
n + n 8 3ln
n
n
+ n 8 = = R (cid:0) (cid:0) (cid:0) (cid:0) 3ln 2 lim n lim n
n 1/ � � �
n 1 a n
n
08.8 1
� +� 8 8 3 � = = = 8 (cid:0) (cid:0) lim n
n ln n 8 2 n
+ n
3
1
n
- (cid:0)
=
n
2
22 n -� � n 1 � �+� � n 2
f x (cid:0) )
+ n
1
3 n
-
n
22 n +� � n 2 � �-� � n 1
= = R (cid:0) (cid:0) (cid:0) (cid:0) lim n lim n
1 a n
+ n
1
3 n
-
= +
(cid:0) (cid:0)
2
1
n
+ n
1
-
1
3 n
3 2 . 1
- -
6e=
=
(cid:0) (cid:0)
n � 3 � �
22 n 3 � � lim 1 �-� � n n � � 3 � + � lim 1 � -� n � n �
1
� n � � �
n
n
2. Tìm miền hội tụ của các chuỗi sau:
(
n
(cid:0) (cid:0) -
) 1
) 1 +
=
n
- a (cid:0) ) b x (cid:0) )
(
0
n
= 1
n +� n 3 � ( �+� � n 2 1 �
n 2 x ) n 5 .3
n
2
n
n
n
+ 1
(cid:0) (cid:0)
(
)
2
=
n 3 n
n
n
0
= 1
+ + c x d (cid:0) (cid:0) ) 2 5 )
n
� 2 � n 3 � � x � �
(
)
(cid:0) -
n
e (cid:0) )
n
= 1
x ( n 8 ) 2 !
Hướng dẫn
n
n
(
(cid:0) -
) 1 +
n
a (cid:0) ) 3R =
(
= 1
n 3
n x ) 5 . 2
(
n
n
- Khoảng hội tụ:
)3,3 )
(
) 1
(cid:0) (cid:0) - -
( +
= x = - 3
n
n
+
� (
� (
)
= 1
= 1
n 1 n 3 ) n 5 .3 2 2 5
(cid:0)
(cid:0) Chuỗi phân kỳ vì cùng bản chất với
1 1/2 n n = 1
n
n
(
(cid:0) -
) 1 +
n
(cid:0)
(
= 1
n
n 2
(
(cid:0) (cid:0) - - x ) n 5 .3 (
) 1 +
n
n
= x = 3
� (
)
� (
= 1
= 1
) n n 1 3 ) + n 5 .3
n n 2 5 2
na
= (cid:0) 0 Chuỗi đan dấu với
+
(
)
1 n 2 5
] 3,3
Chuỗi ht theo tc Leibnitz. ( MHT D = - :
n
n
(cid:0)
) 1
n
= 1
b (cid:0) ) 2R =
+� n 3 � - ( x �+� � n 2 1 �
) = -
(
)
- Khoảng hội tụ: ( + 1 2,1 2 1,3
x = - 1
n =
(cid:0) (cid:0) (cid:0)
)
) n = 1
- - 2
� a n
n
n
n
= 1
= 1
= 1
n � ( � �
n � ( � �
+ + + 3 + 1 6 1 n � � � n 2 � n 2 � � � n 2 �
n =
(cid:0) (cid:0) (cid:0)
)
) n = 1
- - 2
� a n
n
n
n
= 1
= 1
= 1
n � ( � �
n � ( � �
+ + + 3 + 1 6 1 n � � � n 2 � n 2 � � � n 2 �
+
n � � = 1 � � � �
n � � �
n
+ 1
5 + n 6 1 2 1 +� n 2 � + n 2 �
n
5/2
(cid:0) (cid:0) = (cid:0) (cid:0) (cid:0) (cid:0) e
n 2 � 5 � �
1
5 . + � n 1 2 � � �
� 5 � + � 1 � +� n � 2 �
na
(cid:0) 0 Chuỗi pk theo đk cần
n
n
(cid:0)
) 1
n
= 1
(cid:0)
+� n 3 � - ( x �+� � n 2 1 �
x = 3
n
(cid:0) (cid:0) (cid:0)
= =
� a n
n
n
n
= 1
= 1
= 1
n � 2 � �
n � � �
+ + + 3 + 1 6 1 n � � � n 2 � n 2 � � � n 2 �
na
)
( MHT D = - :
(cid:0) 0 Chuỗi pk theo đk cần
1,3
2
n
n
(cid:0)
(
)
=
n
0
+ c x (cid:0) ) 2 5 0R =
Chuỗi chỉ hội tụ tại: x = - 5
n
n
2
n
+ 1
n 2 .
+ 1
(cid:0) (cid:0)
2
2
n
= 1
= d (cid:0) ) (cid:0)
n
= 1
n �+ 3 x � n �
n n 3 . � 2 � n 3 � � � � �+ 9 n x � n �
n
R =
n
2
n 2 .
(cid:0)
2
x = - (cid:0)
n
= 1
1 3 1 3 n n 3 .
(cid:0)
n
= 1
= - (cid:0)
+ � �- 1 9 � � n 3 � � ) ( n �- 1 � 2 n � � � n 2 � � +� � � 9 � �� �
HT HT HT
n
n
2
n 2 .
(cid:0)
2
9 x = (cid:0)
n
= 1
1 3 n n 3 . + � � 1 � � n 3 � �
(cid:0)
= (cid:0)
n
= 1
n � 2 � � +� � � 9 � �� �
1 2 n � � � �
HT HT HT
= - MHT D :
1 1 , 3 3 � � � � � �
n
(
)
(cid:0) -
n
e R = +(cid:0) (cid:0) )
n
= 1
)
( MHT D = - :
x ( n 8 ) 2 !
(cid:0) +(cid:0) ,
3. Tìm khai triển Maclaurin của các hàm số sau:
2
)
)
( a f x )
( b f x )
x 2 = = x sin
(
) 2
- x 1
)
(
)
)
( c f x )
( ln 1 2
= - - x x 2
)
( d f x )
=
x 2 + x 3
2
(
)
Hướng dẫn x
= x fa ) s in
)
(
= - x 1 cos 2
n
2
1 2
(
2
(cid:0)
(
) 1
=
) x 2 ( ) n 2 !
n
0
= - (cid:0)
� 1 1 -� � 2 2 � � � � �
(
)
x 2 = x fb )
) 2
(
-
)
(
) (
) (
)
2
3
)
(
(
)
(
)
- - - - - x ) ( 1 ( 2 3 2 4 = + - - - - L + x x + x x 2
n
2
3
2! 3 3! � � �
(
= + + + + + + +
)
) 1
L L x x x x n x � 1 2 � � ( 2 1 2 3 4
( -�
)1,1
x- ĐKKT:
)
(
)
)
( c f x )
( ln 1 2
n
= - - x x 2
(
)
n
1
(
)
(
) 1
(cid:0) - - (cid:0) = - - - < - 1 2 1x x (cid:0) 2
n
= 1
x 2 n
n
n
n
+ 1
+ 1
(cid:0)
= + n -
(
)
x x (cid:0) 2 2
n
= 1
1 n
+ 1
n
n
1
n
(cid:0) - (cid:0) -
+ x = (cid:0) (cid:0)
=
n
= 1
n
2
- 2n n 2 n x 1
4. Tìm khai triển Taylor của các hàm số sau:
)
( a f x )
0
1 = = x , 3
- x 1
p
)
( b f x )
= = x x sin ,
2
p p
)
( c f x )
= - x arctan
4 � � = x , � � 4 � �
4. Tính tổng của các chuỗi lũy thừa sau:
n
(cid:0)
( -�
) 1,1
x (cid:0) 1) ,
)
( n n
n
= 1
+ + n 2 x ) ( 1
n
1
)
- (cid:0)
(cid:0) 2)
( n x n
n
= 1
+ + ( 3 1)!
Hướng dẫn
n
( -�
(cid:0)
) 1 1,
(cid:0) ) 1 x ,
)
( n n
n
= 1
+ + n 2 x ) ( 1
n
(cid:0)
)
( S x
= - (cid:0) .
n
= 1
1 + 1 + n n n 1 + 1 2 2 1 1 � . � 2 � � x � �
n
n
n
(cid:0) (cid:0) (cid:0)
= -
n
n
n
= 1
= 1
= 1
+
x x � � + n n x � + n 1 2 1 + 1 2 2
n
+ 1
2
(cid:0) (cid:0)
)
( ln 1
( - � x ,
} { ) 1,1 \ 0
= - - - x
n x + � + n
n
n
= 1
= 1
+ 1 x 1 1 . 2 x x � n 1 2 2 1 2
+
n
+ 1
2
(cid:0) (cid:0)
)
( ln 1
( - � x ,
) } { 1,1 \ 0
= - - - x
n x + � + n
n
n
= 1
= 1
+ 1 x 1 1 . 2 x x � n 1 2 2 1 2
n
n
(cid:0) (cid:0)
)
{
( - � x
( ln 1
} ) 1,1 \ 0
2
=
=
= - - - x ,
n
n
2
2
1 x x + � n x � n 1 x 2 1 2
)
)
( ln 1
( ln 1
2
= - - - - - - x x x � � � � 1 x 1 2
)
{
( ln 1
} ) 1,1 \ 0
2
+ - - - - x x
1 x 2 x 2 � � � � ( - � x , � �
)
) + - x
( S x
( - � x ,
} { ) 1,1 \ 0
2
= - -
1 x 2 3 4 1 x 2 1 1 � - + � x 2 � � ( ln 1 � �
= x 0
n
(cid:0)
(
)
= = (cid:0) S 0 0
)
( n n
n
= 1
+ + n 2 0 ) ( 1
n
1
)
- (cid:0)
( n x n
(cid:0) 2) MHT D R= :
n
= 1
( 3 1)!
n
1
)
)
( S x
- (cid:0) + - + + ( n + x 3 = (cid:0)
n
= 1
1)!
n
n
1
1
) ( 1 1 + n ( )
)
- - (cid:0) (cid:0) + x
( �
( �
- =
n
n
= 1
= 1
n
n
+ 1
x n 3 n ! + 3 + 1)! (
)
)
(cid:0) (cid:0) + x
2
( �
= -
(
)
n
n
= 1
= 1
1 + 1 + x n x ( n 3 3 ! + 3 ) + 1 ! x 3
( � x (cid:0)
- 3
n
n
+ 1
)
)
(cid:0) (cid:0) + x
2
( �
( �
= -
(
)
n
n
= 1
= 1
n
1 + 1 + x n x ( n 3 3 ! + 3 ) + 1 ! x 3
(
)
+
x
3
(cid:0) + x
= - -
(
e (cid:0)
) 1
2
=
(
)
n
2
+
+
x
x
3
3
1 + 1 + n x 3 ! 3 x 3
= - - - - - -
(
(
)
e e x
) 1
2
1 3 x (cid:0) 3
(
)
1 + 1 + x 3 x 3
n
1
(
- (cid:0)
) - = 3
(
(cid:0) S
n
= 1
n .0 + n = ) 1 ! 1 2
4. Tính tổng của các chuỗi số sau:
(cid:0) (cid:0) -
(
n
= 1
n
) n 3 (
= 1
(cid:0) 1) (cid:0) 2) n n - - 4 3 ( ) 1 + n 1)! 7
(cid:0) (cid:0)
1 (cid:0) 4) (cid:0) 3) 1 n
(
n
(
= 1
) 3 (2 )!!
n
= 1
) n 3 (2
- - n + n 1)!
(cid:0)
(cid:0) 5)
)
( n n n = 1
+ + n 2 1 ) ( 1
(cid:0)
n
(cid:0) 1)
(
)
n
= 1
n
n
+ 1
- 1 + n 3 ) 1 !
)
(cid:0) (cid:0) - - ( )
( � 3
= = -
( � (
n
n
= 1
= 1
1/ 3 + n 1)! ( 1)!
)
n = -
=
(cid:0) - 1/ 3 + n ( - = - - + 1/3 (cid:0) (cid:0) (cid:0) 3 3 . 1
n
2
1/ 3 n ! 1 3 � e � � �
(cid:0) (cid:0) -
(
= - - (cid:0) (cid:0) 2) + n 3 3 n n
n
n
= 1
= 1
n 1 � � ) - � � 7 7 � �
- 4 3 ( ) 7
(cid:0) (cid:0)
(
) 1
n
n
= 1
= 1
n -� � 1 + � � 7 � �
n 1 � � � � 7 � �
= - - + n (cid:0) (cid:0) 3 7
-
(
)
= - - S 3 + 1/ 7 7.
Trong đó S(-1/7) tương ứng với
1/ 7 + 1 1/ 7
(cid:0)
)
(
( S x
) 1 n x
n
= 1
= + (cid:0) n
Tính S(x)
x
(cid:0)
n
+ 1
( -�
) ( S t dt
) 1,1
0
n
= 1
x = = (cid:0) x x x . , (cid:0) - x 1
2
2
(cid:0)
)
( �
( S x
) 1,1
2
= - � x ,
- 2 ( x ) x 1 � � - x x = � �- x 1 � �
(
(
)
) 2. 1/ 7 ( +
)
2 7 - = - 8
n
= 1
(cid:0) - - - - 7 = - (cid:0) � 3 n n 1/ 2 - 4 3 ( ) 11 64 7 1 1/ 7
(cid:0)
1
(cid:0) 3)
(
n
= 1
) n 3 (2
+ 1
+ - n 1) !
(cid:0) (cid:0)
) n 1 .
) n 1 .
( �
( �
= - - 3
n
n
= 1
= 1
n
+ 1
2
1
+ 2
n 2 � � 1 � � 3 � � = ( ) + n 1 ! 2 (
n 2 � � 1 � � 3 � � ( ) n 1 ! ) (
0
(cid:0) 1/ 3
(
(
) n 1 .
) 1
= - - - (cid:0) 3
n
) 1/ 3 ( ) + 1 !
0
2 1!
� � � � � � � � � = n �
= -
1 3 1 3 � 3 sin � � � � �
(cid:0)
(cid:0) 4) 1 n
(
n
= 1
) 3 (2 )!!
- n
(cid:0)
n
n
= 1
n
= (cid:0) 1 n -
( (
) n 3 .2 . ! )
1/6
(cid:0) - - = = - (cid:0) e 1
n
= 1
1/ 6 n !
(cid:0)
(cid:0) 5)
)
( n n n = 1
+ + n 2 1 ) ( 1
Không dùng chuỗi lũy thừa, chỉ qua giới hạn của dãy tổng riêng phần Sn
= -
)
( n n
+ + 1 + n n n 1 1 + n 1 ) ( 1 2 1 1 2 + 1 2 2
= + + + ... S n a n a 1 a 2
nS
= ...
1 1 2 k 1 n 1 + + + + 3 1 2 1 2 � � �
- -
1 1 + + + + ... n n 2 1 3 � 1 � � � � � 1 � �+ 1 �
+ + + + + ...
1 + 1 + 1 n n n 1 + 1k 1 1 2 2k + 1 2 1 3 1 2 � � � � � �
= + -
1 + 1 + 1 + n n 1 2 1 + 1 2 1 1 � � � + 1 � � � 2 2 � � � 1 � � + � � n 2 � � � � �
n(cid:0)
(cid:0) (cid:0) (cid:0) (cid:0) (cid:0)
1 2 1 - = 2 1 4 1 � �+ 1 � � 2 � �
