3???1?例6:?; n3n?1?nn2n3???1?4nnlim2?lim4?1, 解:由于,及??nn??n??333nn3???1?1limnun?limn??1故级数收敛. nn??n??33n例7:
1?p?0? ?pn?2nlnn????lnlnx???,???2dx?解:由于I????2??xlnpx?1?lnx?1?p,?2?1?pp?1,及当p?1时,I?p?1;1?ln2?1?p, p?1当0?p?1时,I???,因此当0?p?1时,级数发散;当p?1时,级数收敛.
例8:
1; ?lnn!n?2??111?解:由于lnn!?ln1?ln2?ln3???lnn?nlnn,即,级数?发散,故lnn!nlnnn?2nlnn原级数也发散. 例9:
n?1??1?ln??; ?n?n?1?n?1n?1?ln?x?ln?1?x?1111nn?x?lim?解:由于lim,取则有,及收敛,故原级数 ?22n??x?0?2nx22n?1n?1????n?n?1??1?ln??收敛. ?n?n?1?n例10:
???lnlnn?n?3?1lnn;
lnn2解:limlnlnn???,当n充分大时,则有lnlnn?e,即?lnlnn?n???e2??lnn?n2,
也即
1?lnlnn?lnn??111?2,又?2收敛,故原级数?收敛. lnnnn?3nn?3?lnlnn?例11:
??lnn?n?1n?nlnnn;
??ln2nne???1,故级数收敛. lim?0解:limun?lim?lim?0??n??n??n??lnnn??lnnn??lnnnln2nn?lnn?例12:??1??;
n?n?1??lnn??1??nlnnn???lnn????n1?解:考虑xn???,lnxn?lnn?nln?1??,
1nn????nn?nlnn1?lnn??lnn??lnn?又,ln?1????????o??,
nn2nn??????2?lnn1?lnn?2?ln2n?n??1ln2n?lnn??lnxn?lnn?n?????o???0, ??o??????n????n2nn2n??????????22即xn????1,
n??1发散,因此原级数收敛. ?n?1n?1?an?q?an?0?,证明级数?an当q?1时收敛,当q?1时发散. 例13:设limn??lnnn?1ln1an证明:根据极限定义,???0,?N?0,当n?N时,恒有?q??.即
lnnln1anq????q??.
lnnln当q?1时,取适当小的?,使q?????1,即得ln11??lnn?lnn?,即an??,
nan?1
又,?????1?收敛,故?an收敛.
n?1nn?1?
当q?1时,取适当小的?,使q?????1,即得ln11??lnn?lnn?,即an??,
nan?1
又,?????1?发散,故?an发散.
n?1nn?1?
例14:已知级数
?a,?b2nn?1n?1??2n收敛,问级数
?abn?1?nn是否收敛?
???121222解:由于anbn?an?bn,及?an?bn收敛,故?anbn收敛,即?anbn绝对收敛,
2n?12n?1n?1????因此
?abn?1?nn收敛.
例15:讨论级数
?n?3???1?nln1nn的收敛性,若收敛,是绝对收敛,还是条件收敛?
????1?lnnlnnlnx12????dx??lnx????知?n3x2n?3n?3nn?1解:原式??n?3???1?n?1lnn,由
n???31lnxlnnlnx?0, 发散,考虑f?x??,由于lim?limx?0,取x?n得limun?limn??n??nx???xx???1x1?lnx?0?x?3?,故f?x?在x?3时单调减少,因此有f?n??f?n?1?,即 又,f??x??x2un?un?1,满足莱布尼兹定理条件,因此本级数条件收敛.
n??1?例16: 讨论级数?nn???1?n?2?的收敛性,若收敛,是绝对收敛,还是条件收敛?
n??1?发散,故?nn???1?nn?2?解:由于
1n???1?n?12n,
?2n?2?1??n?2?1n???1?nn发散,
又,
??1?nnn???1?n??1???????1n???1??1?n1?1?n,级数?收敛,???n?1n?1n?1n?1n?2n?1n?2n?nn??1?发散,故原级数?nn???1?n?2?发散.
例17: 讨论级数
?sin(n??lnn)的收敛性,若收敛,是绝对收敛,还是条件收敛?
n?2?1解:由于sin?n????11?1n?0.故原级数为交错级数. ,且当n?2时,sin????1?sinlnnlnn?lnn1???111nlnn发散(,又发散),故原级数不会绝对???1sin?sinlim?1???sinn?2lnnn?2lnnn??1n?2lnnlnn收敛,又,limn??u1n?limn??sinlnn?0, 取f?x??sin1lnx,f??x??cos1?1?1lnx???ln2x??x?0?x?2?,故f?x?单调f?n??f?n?1?,也即un?un?1,满足狄里克来条件,因此该级数条件收敛.
??例18:已知数列?nan?收敛,级数
?n?an?an?1?收敛,证明级数n?1?an收敛.
n?1?n??解一:设级数
?an的部分和为Sn?n?an?an?1?部n?1?ak,级数
k?1n?1?nSn??k?ak?ak?1??a1?a0?2?a2?a1??3?a3?a2????n?an?an?1? k?1 ?nan?(a1?a2???an?1)?a0?nan?Sn?a0
??即limn??Sn?limn???nan?S?n?a0?存在,故级数?an收敛级数
n?1?an收敛.
n?1?解二:取Sn?nan,则级数
??nan??n?1?an?1?收敛,又,?nan??n?1?an?1?n?ann?1???an?1,故级数?an?1收敛,因此原级数?an收敛.
n?1n?1例19??:已知?cn??c证明级数????c?n?0?收敛,?1??nk?1k2?n2?也收敛.
n?1nn??证明: 已知?cn??ccn?0?收敛,故存在M?0,使?1n?n?M,.
nn?1n少,即和为
an?1??
减分?记Ap,qpcncn???2??2n?1k?1k?nn?1npqn1dx11?1,由于, lim????222?0n??41?xk?1k?1?k?n?k?n1???1????n??n?q11???cn?故??,因此Ap,q???M,即正项级数的部分和数列?Ap,q?有上24n?1n4k?1?k?n41????n?1c???界,从而原级数???2n2?收敛.
n?1?k?1k?n??q?1?n例20?1998,2?:设正项数列?an?单调减少,且级数???1?an发散,试问级数???a?1??是
n?1?nn?1???n否收敛?并说明理由.
解: 正项数列?an?单调减少且有下界?an?0?,故liman存在,不妨设为a,则a?0.若a?0,
n??则由莱布尼兹定理知级数
n???1?n?1?nan收敛,矛盾,故a?0.因此由
n??1??1?1n???,故级数收敛. limun?limn???1???n??n???a?1?a?1n?1?an?1??n??n1?n?1?1?例21?2002是否收,1?:设un?0?n?1,2,??且lim?1,问级数???1????u?n??un?1n?nun?1?敛? 解:Sn?k?1???1???k?1n?11??? ??ukuk?1?=
1111111?1n?11n?1?1???, ???1??????????1?????uuu1u2u2u3u3u41n?1?unun?1?1n1?1知lim?0,于是limSn?,级数收敛.
n??un??un??u1nn由lim11?unun?1nn?1n?lim???2故又,limn??n??u1un?1n?1nnn?1???1???n?1??n?1?11?发散,因此原级数unun?1?11??条件收敛. ???unun?1?