(91)-高数专项练题11_08.2026考研数学高途王喆全程班_赠送2025课程_25考研数学（三）全年智达班_{2}--资料

 
例7.10 若级数  4na 条件收敛，则  a ( x  1)2n 的收敛半径
n n
n1 n0
R 为( )
C
(Ａ)R  3 (Ｂ)R  4 (Ｃ)R  2 (Ｄ)无法确定R
-an(4EFUSE
:
anX" 4
Rp
=
Xanthan t "
Gutt=
R 4
: =
R 2
=
. . 
例7.11 设幂级数  a (x  1)n 在 x  1处条件收敛,则  a ( ).
C
n n
n1 n1
(A)发散 (B)条件收敛 (C)绝对收敛 (D)无法确定敛散性
:
an UNBER
& EIEFUR = 2
an
an SSEJYN/
2)
(2
.
: an S
=
an
,  5 1
例7.12 设幂级数  a xn 与  b xn 的收敛半径分别为 和 ，则幂级
n n
3 3
n1 n1
 a 2
数  n x n 的收敛半径为 ( )
A
2
b
n1 n
5 1 1
(A) 5. (B) . (C) . (D) .
3 3 5
= =Matt
R P
:
,
5
P um bl
- R2 = : = 3 =
Ma
=Ma
·
+
R
= = = 5 n 2  1
例7.13 求幂级数  (2x  1) n1 的收敛域及和函数.
n
n1
& E2 X
-
C
+ n
=
U
P
n+1
A
Eter
2/
:U ( 1)
te
.
+
=
/Sl tE( 1)
,
Sult
S (t) +
= .== (i)
citi
Sil
=
=
=
t t
= +St
Salt)
,
-e
(tSuti' +
=
:
= 1
to in
Cusicula
one
=
:
-EInU
t)
-In (1 t) : EtFOAl Salt) =
tSzlt
=> - =
=
u-e-Em(t)
te
SM = 9 or Call
: ,
2,
t
= 0t It
2 X+
=
-I
= xl
Sil
&
5
x
=
2
I
-
UEXE (0 1)
.xn1 
例7.14 设u (x)  enx  （n  1,2, ），求级数  u (x)的收敛域
n n
n(n  1)
n1
及和函数.
=
Uni
+e *
z e E ke
+ - ) = x10
=
e
EX0Af :
, 1-eX
=
Si
unitas
Pho ni
1 IAF U2E
= 1 = = X =
.:Xu
TBURR**
: E 10 1]
.
.
= nini * Si == x
Sal
lo'
#Situat at
=
Si(x In Cr-X)
= -
1.
1 Sitat InG-t)dt
= -
1 * 1-t 1
+
:
Si
=
--In(t)+ dE
1- t
.
= - X((i -X) + x + (n(1 - X)ScM = - X(( -X) + x + (n(1 - X)
InC-N)
C X) +X
= - ·
exx
TB SM (1 x) m(rx +X xt(0 1)
= = + - . , ,
I
et
X /
=
1
+
et
1 -1
 
例7.15 设数列 a 满足a  1，(n  1)a  (n  )a ，证当 x  1时级
n 1 n1 n
2

数  a xn 收敛，并求和函数.
n
n1
(n E) An
Cu+ ) An
=
+
·
an
=
M
·P
=
11 Ca U
Y
=
=
/ anxi
SIN
KT@
Six = anyt n= X" X
(kTkt = (n + ) an
- . +
ko
k
=
n+
+ X +
(u+
1
X
a = -
=+ AnX +
Cut C
SiN 1
Gi · =
=
+
Han an
=
&
Six SIM
= It X . +
-b) S 1x
ES Six
(i-X) Six 1
=> - = = - =
·
24
Izaax
(((2
my
xax
= SN = e . + c
e-E(X)
)SeE(X) y
.
= ax C
+
/my ax
+
c
= +/my ax
+
c
= +
1
4
( 2
= . - x +
1- X
E
2
-
=
F
Siol = C = 2
& = o
=
: Si -2 1
例7.16
求数项级数

的和
(n2

1)2n
n2
t =
·x"
=nit
Six
73
UEEE [ 1 1]
.
1 x ( ul x)
2 -
= -
SIN
=
Curl (n-1)
-
=
=
Sit-2XSI)
=S
=
Si)
1 -X
lo1-Eat Sin -In(-x)
l =
Sitiat =
=
=
=
= n, Six X
Sil
++ (
( at 7 + 1 -j)at
#Scat = - +
=
1 - t ,
E
(n( -x)
X
Sa = - - - (-1 0)
,
1)
10
.
* Ex)
* MG-X mu x))x
:
Six
=
ESN- Su =&- - - - x - - = 0
O X
=0
x=/
T S(2) E E1n2
:
= -
=-
 n  1
例7.17  (1) n ( A )
(2n  1)!
n0
1
(A) (sin1  cos1) (B)2sin1 cos1 (C)sin1 2cos1 (D)sin1 cos1
2
= Eu
cosX =
ShX
I = Hi + Curtis /
TB
: !
!(10s) sm)
+
= xf (x)
例7.18 设 f ( x)   xn,则F(x)  展开为
1  x
n0
x 的幂级数为________.
=
fi
i =
: F = =
x(x
-x)
( -y x(
=
= =
,
(nx) (t) =
x
X
= X = .
.
xt(- 1)
,1  x2
 arctan x, x  0,
例7.19 设 f (x)   x 试将 f (x)展开成

1, x  0.

x 的幂级数，
 (1)n

并求级数 的和.
1  4n2
n1
Ex
ExY" = xi
Carimx)
=
=
=
+ 1 -
= Ell"
that int Youth
animX
=
fix . A
(
X)
- +
=
EfIN =
:
m
It
=
It Citin
=
H
I r xE 1
= = ,
,
(2
1
X=
Fin
2
fill == I
=> +
=
-