21．解: （1）an?1?2(n?1)an2an?n1 ??2an?nan?12(n?1)an?n?12an?nn?11nn?11n?????1??2?(?2), an?12anan?12anan?12ann11n?11n11?2n?1n?2n ??2?(?2)()?n??2?n??an?ana122an22n1?2n?1anan?12n111（2）由bn???(?) n?1n?1n?2n?1n?2n(n?1)2(1?2)(1?2)21?21?2知Tn?1111(?)limT?，所以. nn??251?2n?210n(n?1)， 222．解：（1）当n?3时，Sn?nan?2?Sn?1?(n?1)an?1?2?(n?1)(n?2)n?1，可得：an?nan?(n?1)an?1??2，
22?an?an?1?1(n?3,n?N?)．?a1?a2?2a2?2?1,?a2?3. ?4,(n?1)可得，an?? ??n?1.(n?2,n?N)（2）1?当n?2时，b2?b12?2?14?3?a2，不等式成立．
2?假设当n?k(k?2,k?N?)时，不等式成立，即bk?k?1.那么，当n?k?1时，
bk?1?bk2?(k?1)bk?2?bk(bk?k?1)?2?2bk?2?2(k?1)?2?2k?k?2,
所以当n?k?1时，不等式也成立.
根据（1?），（2?）可知，当n?2,n?N?时，bn?an.
1?x?1??0, 1?x1?x?f(x)在(0,??)上单调递减，?f(x)?f(0),?1n(1?x)?x.
111?, ∵当n?2,n?N?时，?bnann?1（3）设f(x)?1n(1?x)?x,f?(x)??ln(1??ln(1??(1?11111)????， bnbn?1bnbn?1(n?1)(n?2)n?1n?21111111111)?1n(1?)???ln(1?)????????? b2b3b3b4bnbn?134n?1n?23n?23111 )(1?)?(1?)?3e.b2b3b3b4bnbn?1