【正文】
)2)(1(1)2(21a r c t a n)1(..212112?????????????nnnnnnnnn;求下列級數(shù)的和30 .43ar c tan,32ar c tan,21ar c tan)1(321 ??? sss解)41ar c tan1ar c tan ?????? nnnsn (所以?.421a r c t a n12?????n n故則假設(shè) ,1ar c tan1 kks k ???,1ar c tan21ar c tan1ar c tan2 ?????kkkkksk31 .41)2)(1(1.41lim)2)(1(12121)2)(1(1)1(121)2)(1(1)2(11????????????????????????????????????????nnnnkknnnnnsnnusnnnnnnnu令32 ? ?? ? .4421a r c t a n)1a r c t a n (l i ma r c t a n)1a r c t a n (11a r c t a n11a r c t a na r c t a n)1a r c t a n (1a r c t a na r c t a na r c t a n)3(1122?????????????????????????????????nnnnnnnnnxyyxyxnnn由33 .,)1(.31102 ??????nnnn udxxxu 求設(shè)? ?.)(.4011收斂級數(shù)證明收斂,收斂,級數(shù)設(shè)數(shù)列????????nnnnnnuuunun34 .61)3121()3121(lim)3121()2111(lim)3121()2111(312211)2()1()1(1110211021102?????????????????????????????????????????????????????????????????nnkkkkunnnnnnndttttdtttdxxxunnknnnnnnnxtnn解35 ? ?..)(limlim)(,)(.lim01110111011收斂級數(shù)故所以且收斂,不妨設(shè)其和為因為級數(shù)又使得存在常數(shù)所以收斂,因為數(shù)列證?????