(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 => + = -