例22?1994,1?:设f?x?在点x?0的某个邻域内具有二阶连续导数,且limn??f?x??0,证明级
x数
?n?1??1?f??绝对收敛. ?n?f?????2f?????2x?x,f???x?在2!2证明:由题意知f?0??f??0??0,又f?x??f?0??f??0?x?点x?0的某个邻域内连续,不妨取该邻域内含有原点的一小区间I,f???x?在I上连续,则
?M?0,使f???x??M,故f?x???1收敛,因此级数??2n?1nn?1?M21?1`?M1x.令x?,当n充分大时,有f???, 22n2n?n??1?f??绝对收敛. ?n?
[例23 ]判别级数
!!n?1?2n?1?的敛散性。 ???1???2n!!n?1?[解]（1）un?2?2n?1?!!??2n?1?!!?2n?1?u，得?u?单调减少；
n?2n?!!?2n?!!2n?2n?12由?2n???2n??1??2n?1??2n?1?，及
?2?4?6??2n??2?1?32?52?72??2n?1?2??2n?1?得
2?4?6??2n??1?3?5??2n?1?2n?1，
1?3?5??2n?1??即0?un?2?4?6??2n??n?112n?1?0?n?0?，故级数???1?n?1?n?1?2n?1?!!收敛。 ?2n?!!?1?3?5??2n?1?1?2n?1?!!??2n?1?!!1?又，???1?， 由，及?发散知 ??2?4?6??2n?2n?2n?!!n?1?2n?!!n?12nn?1??!!2n?1?!!n?1?2n?1???发散，故级数条件收敛。 ?1???1???2n?!!?2n?!!n?1n?1?n?1
一、 函数项级数 1. 概念与性质
(1)
?u?x??u?x??u?x??u?x????u?x???称为函数项级数.
n123nn?1?若
?u?x?收敛,称xn0n?1?n?0为
?u?x?的收敛点,收敛点的全体称为收敛域, 若?u?x?发
nn1n?1n?1??散,称x1为
?u?x?的发散点,发散点的全体称为发散域.
n?1若x为收敛点,
?u?x?nn?1?收敛于和函数S?x?,记Sn?x????u?x?kk?1n,则
rn?x??S?x??Sn?x?称为级数?un?x?的余项,并且limrn?x??0.
n?1n??(2)记u0(x)?a0,un?x??an?x?x0?,n?1,2,?则
n?u?x??u?x???u?x??a??a?x?x???a?x?x?nn0n0n0n0n?0n?1n?1n?0????n称为关于x?x0的幂级数,记t?x?x0,则有
??atnn?0?n,为了方便以后记为
?an?0?nxn.
(3)Abel定理:如果幂级数
??an?0n在x?x0?x0?0?处收敛,则对于适合x?x0的一切xnx,?anx绝对收敛, 如果幂级数?anxn在x?x1处发散,则对于适合x?x1的一切
nn?0?n?0?x,?anxn发散.
n?0(4)若幂级数
?an?0?nxn在某些点收敛, 在某些点发散,则必存在某个确定的唯一的常数R,
??使当x?R时,幂级数
?an?0nx绝对收敛;当x?R时,幂级数?anxn发散.这里的R称
nn?0为幂级数
?an?0?nxn的收敛半径.
?(5) 收敛半径的求法:设an,an?1为幂级数
?anxn的相邻两项的系数,若limn?0an?1??或
n??anlimnan??,则当??0时,取R?n??1?;当??0时,取R???;当????时,取R?0
注意:R?0时, 幂级数2. 幂级数的运算性质
(1) 若
?an?0?n仅在x?0处收敛. xn?a?nx在??a,a?内收敛,
n?bxn?n在??b,b?内收敛,则
??a?n?bn?xn
n?0n?0n?0????在??c,c?内收敛,其中c?min?a,b?且
?annnn?bn?x?nx?n?0?a?bnx.
n?0n?0??????n?an??n?xnnx???bnx??n?0??cn,其中cn?n?0?akbn?k,n?1,2,?
n?0??k?0?(2)
?annx的和函数S?x?在其收敛区间内连续,可导,且可逐项求导和逐项积分n?0?? S??x???????an???n???nx??n?0??anxn?0?nanxn?1. n?1
?xS?x?dx??x?n?xn?an?1nx00(?anx)dx?n?0?(n?0?0anxdx)??n?0n?1 注意: 逐项求导和逐项积分后得到的幂级数与原来的幂级数具有相同的收敛半径区间端点的收敛性要重新讨论. 3. 泰勒级数
??n?(1) 若f?x?在x0的某个邻域U?x0?内具有各阶导数,则
?f?x0?!(x?xnn?0n0)x?x0的泰勒级数.
(2) 若f?x?在x0的某个邻域U?x0?内具有各阶导数,则f?x?在该邻域内能展开成泰勒
级数的充分必要条件是limn??Rn?x??0 ,
?x??f?n?1?其中R???n?n?1?!?x?x?10?n, ??x0???x?x0?,0???1. 注意: f?x?在该邻域内能展开成泰勒级数是指f?x?的泰勒级数在该邻域内收敛敛于f?x?本身.
直接展开法:
(1) 求出f?x?的各阶导数:f?n??x?,n?1,2,?及f?n??x0?,n?1,2,?,
,即,, 但是
称为关于且收(2) 写出泰勒级数:
?n?0?f?n??x0??x?x0?n,并且求出它的收敛半径R, n!f?n?1????(3) 考虑当x???R,R?时,Rn?x???x?x0?n?1(其中?介于x,x0之间)在n?0时,?n?1?!是否趋于零,如果趋于零,则有f?x????f?n??x0??x?x0?,即f?x?可展开成泰勒级数. n?0n!间接展开法(即利用常用的函数展开式):
(1)1?1?x??xn,x?1; n?0(2)ex???xn,x?n?0n!???,???;
(3)sinx?????1?n?1x2n?1?2n?1?!,x????,???;
n?1(4)?1?x???1??x?????1?22!x????????1????2?????n?1?n!xn??
x???1,1?,端点的收敛性与?有关.
4、举例
例1:求下列幂级数的收敛域 ?(1)
?3nn?1nxn;
3n?1解:??liman?111?nn??a?limn?1nn??3n?3 ,所以R?3.当x??3时,???1?为交错级数n?1nn莱布尼兹定理要求,故该级数收敛,当x?1?113时,?为p?级数, p?n?1n2?1,故级数发散因此该级数收敛域为???1?3,1?3??. (2)
????1?n2nx2n?1; n?1,满足,
nun?1?x???1?2n?1x2n?12?lim?2x解:该级数为缺项级数,直接用比值法判别:lim
n??u?x?n????1?n?12nx2n?1n当2x2?1,即x?12时,级数收敛,当x?12时,un?x??2nx2n?1?1??2n??2??2n?1?2即limun?x??0,也即limun?x??0.因此该级数收敛域为??n??n????12,1??. 2?1?1?x?(3):???:
2n?11?x??n?1?a1?x1n2n?3解:取t?,则有?t,由??limn?1?lim?1知R?1
n??n??1?xan2n?1n?12n?1?1?x1?x1?1时,x?0,当x?0时,?当发散,故该级数收敛域为?1,即,?1?1?x1?xn?12n?1?n?0,???.
3n???2?(4)??x?1?n
nn?1?n解: 由??liman?1n??an3n?1???2??limnn?1nn??3???2?n?n?1?3知R?1 314当x?1??,即x??时,
33由于
3n???2??nn?1nnnn1?2???1?????1?????(*) ???=??n?3???3?n?1??n??n?1???1?n,
n1?2???收敛,故(*)收敛. ?n?1n?3?nnnn??1??1??2??3n???2??1????(**) ??=????n?3??n?3?n?1?n?1?n??n12当x?1?,即x??时,
33???1??1??2?收敛,故(**)发散. 故该级数收敛域为??4,?2?.
由于?发散,????33?nn3????n?1n?1nn?(5):
?nn?21n(p为常数) xplnn