§1.2 stolz定理及其应用
定理1 设{an}是趋于零的数列, {bn}严格递减趋于零,则当liman?1?anbn?1?bn存在或为??、
n?????时，有limn???anbn?limn???an?1?anbn?1?bn.
证明 设limn???an?1?anbn?1?bn?l.
(1) 若l是有限实数,则???0,?N?0,当n?N时,有l???an?1?anbn?1?bn?l??.
由于bn?1?bn?0,所以
?l?l???(bn?1?bn)?an?1?an??l???(bn?1?bn), ???(bn?2?bn?1)?an?2?an?1??l???(bn?2?bn?1),
………
?l???(bn?p?bn?p?1)?an?p?an?p?1??l???(bn?p?bn?p?1),
上述各式相加得 ?l???(bn?p?bn)?an?p?an??l???(bn?p?bn). 在上式中固定n并令p???,由于an?p?0,bn?p?0,得 ?l???bn?an??l???bn.
.所以limn???注意到bn?0,由上式便得
anbn?l??anbn?l.
(2)若l???,则?K?0,?N?0,当n?N时,有
an?1?anbn?1?bn?K.仿照(1)中的证法可得,对任
意自然数p,有
an?p?anbn?p?bn?K,固定n并令p???,得
anbn?K.所以limn???anbn???.
(3)若l???,可用?an代替an转化为(2)的情形.
?定理2 设{an}是任意数列, {bn}严格递增趋于?,则当limn???an?1?anbn?1?bn存在或为??、??时，有limn???anbn?limn???an?1?anbn?1?bn.
证明 设limn???an?1?anbn?1?bn?l.
(1) 若l是有限实数,则??由于bn?1?bn?0,所以??l??0,?N?0,当n?N时,有l??2?an?1?anbn?1?bn?l??2.
??????(bn?1?bn)?an?1?an??l?22????(bn?1?bn)?,
,
???????l??(bn?2?bn?1)?an?2?an?1??l??(bn?2?bn?1)2?2???………
???????l??(bn?p?bn?p?1)?an?p?an?p?1??l??(bn?p?bn?p?1),
2?2???上述各式相加得
???l?2?????(bn?p?bn)?an?p?an??l?2??an?p?anbn?p?bn?l???(bn?p?bn)?.
由此便得 l??2??2.
所以
an?p?anbn?p?bn?l??2.
由恒等式
anbn?l?aN?lbNbn?bN??1??bn??????an?aN???l??b?b?N?n?
得
anbn?l?aN?lbNbn?an?aNbn?bN?l
由于bn???(n???),?N1?0,当n?N1时,有
aN?lbNbn??2.
因此当n?max{N,N1}时,
anbn?l??2??2??.这证明了limn???anbn?l.
(2)若l???,则当n充分大时,有an?1?an?bn?1?bn.由bn???(n???),
bn?1?bnan?1?an可知an???(n???),且数列{an}严格递增.注意到lim?0,
n???由(1)的结论得limn???bnan?0.从而limn???anbn???.
(3)若l???,可用?an代替an转化为(2)的情形.
定理1与定理2统一称为Stolz定理.
例1 利用Stolz定理.证明(§1例7)：设liman?a.证明limn???a1?2a2???nan?2n?a2.
n???证明 令An?a1?2a2???nan, BnAnBnAn?1?AnBn?1?Bn?n2,则{Bn}严格递增趋于??limn???,由定理2,
a2limn????limn????limn???(n?1)an?1(n?1)2n?1?n?1??n22an?1?12n???liman?．
例2 求极限limn???1k?2k???nk?1k,其中k为自然数.
nk解 令ananbn?1k?2???nk, bn?nk?1,由定理2,
(n?1)kk?1limn????limn???an?1?anbn?1?bn?limn????n?1?k?1?limn???(n?1)k?n?k?1?nk???1k?1.
其中倒数第二式中…表示关于n的次数为k 例
?1的一个多项式.
?1k?2k???nk1?3 求极限limn???k?1??n???k?1n????k?1?(1k,其中k为自然数.
解 令ananbn?2k???n)?nkk?1, bn??k?1?nk,由定理2,
klimn????limn???an?1?anbn?1?bn?limn????k?1?(n?1)?nk?1k??n?1?kk?1?k12?1?[?n?1??n]
?limn????k?1?k??k?1?k??2??k?1?knk?1?k?1n???????.
?2的多项式.
其中倒数第二式分子与分母中的…均表示关于n的次数为k注 例3中当k不是自然数时,只要k1n?0(该条件保证
limn????k?1?nk???),利用定理2,并令
x?,我们有limn???anbn?limn???an?1?anbn?1?bn
?limn????k?1?(n?1)k?nk?1k??n?1?kk?1?k?1?[?n?1??n]?limn???1??k?1??n??k1???n?n?1??n??kk
?k??1??1???1??n????k??1???k?1?x???limx?0k1x?1xk?1?x?k?k?1??1?x??1???limx?0kx?1?x??1??1?x?k?k?1?x?1?x??x?k?.
12再利用求函数极限的罗必塔法则,可以求出最后一式的极限为.
例4设limn(An?An?1)?0.试证:极限limn???A1?A2???Annn???存在时,limAn?limn???A1?A2???Ann.
n???证明 因An?An?而极限limn???A1?A2???Ann?A1?A2???Ann,
A1?A2???Ann存在,故只需证明第一项趋于零.
令a1?A1,a2?A2?A1,…,an?An?An?1,则由条件limn(An?An?1)?0知limnan?0,
n???n???且An?(An?An?1)?(An?1?An?2)?A1?A2???An?于是lim??An??
n?????(A2?A1)?A1?an?an?1???a1.
?n?a1??a1?a2?????a1?a2???an????lim??a1?a2???an???n???n???limn???
a2?2a3????n?1?ann(应用定理2)
?limn???(n?1)ann??n?1??limn???n?1n?n?an?0.
例5 设0?x1?1,xn?1?xn?1?xn?(n?1,2,?).证明limnxn?1.
n???证明 由条件
xn?1xn?1?xn.用数学归纳法容易证明对所有自然数
n有0?xn?1,即
存在,
0?1?xn?1.所以数列{xn}是严格单调递减有下界的.由单调有界定理,极限limxnn???设极限值为a.在xn?1?xn?1?xn?中令n???得a由于{1xn}严格单调递增趋于???a?1?a?,由此得a?0.
,根据定理2, limnxn?limn???n1xn
n????limn????n1?1??n?1xn?limn???xnxn?1xn?xn?1?limn???xnxn?1x2n?lim(1?xn)?1.
n???xn?1§1.3 利用压缩影像原理和单调有界定理求极限
压缩影像原理 设f(x)可导且f'?x??r?1,r是常数.给定x0,令xn?f(xn?1)(n?1,2,?).证明序列{xn}收敛.
证明 由拉格朗日中值定理，得
xn?1?xn?f(xn)?f(xn?1)?f'???(xn?xn?1)?rxn?xn?1?r2xn?1?xn?2???rnx1?x0
其中?介于xn,xn?1之间.故对任意自然数p有
xn?p?xn?xn?p?xn?p?1?xn?p?1?xn?p?2???xn?1?xn?rn?p?1
x1?x0?r1?rpn?p?2x1?x0???rrnnx1?x0?rnx1?x01?r???rr?1).
?p?1??rnx1?x0?1?r?x1?x0?1?r?0（n???,0?由柯西收敛准则{xn}收敛.
注 (1)利用压缩影像原理必须保证{xn}是否保持在f'?x??r?1成立的范围之内. (2) f(x)称为压缩映射(因为0?例1 设x1?0,xn?1?解 令f?x??又
f'?x??3?1?x?3?xr?1)．
3?1?xn3?xn?(n?1,2,?),求极限limxn.
n???(x23?0),则xn?1?f(xn)(n?1,2,?). ),故f(x)称为压缩映射.
3?1?xn3?xn6(3?x)2?(x?0由压缩影像原理,{xn}收敛.再对递推公式xn?1?
?,两边取极限即可.
例2设K是正数,x0?0,对任意自然数n,令xn?证明 令f?x??(x?K1?K?xn?1?2?xn?1???.证明limxn??n????f'?x??K.
1?K??x??2?x?(x?12K),则xn?f(xn?1)(n?1,2,?).又
1?K??1?2? 2?x?),从而有0?f'?x??.故f(x)称为压缩映射.由压缩影像原理, {xn}收敛.再对递
推公式xn?1?K?xn?1?2?xn?1???,两边取极限即可. ??