(1)证明:由cosan-an=cosbn,及0<an<π2,0<bn<π2可得0<an=cosan−cosbn<π2,所以0<an<bn<π2,由于级数∞n=1bn收敛,所以级数∞n=1an也收敛,由收敛的必要条件可得limn→∞an=0.(2)证明:由于0... 结果一 题目 设数列{an},{bn}满足0<an<π2,0<bn<π2,cosan−an=cosbn且级数∑n...
cosan−cosbnbn=2sinan+bn2sinbn−an2bn≤2an+bn2bn−an2bn=b2n−a2n2bn<b2n2bn=bn2由于级数∞n=1bn收敛,由正项级数的比较审敛法可知级数∞n=1anbn收敛. (1)由题目条件可知,cosan-cosbn=an>0,根据函数cosx在[0,π2]上单调递减,可知0<an<bn,由于正项级数∞n=1bn收敛可得∞n=1an...
cosbn＜π2，所以0＜an＜bn＜π2，由于级数∞n＝1bn收敛，所以级数∞n＝1an也收敛，由收敛的必要条件可得limn→∞an＝0．（2）证明：由于0＜an＜π2，0＜bn＜π2，所以sinan+bn2≤an+bn2，sinbn?an2≤bn?an2anbn＝cosan?cosbnbn＝2sinan+bn2sinbn?an2bn ≤2an+bn2bn?an2bn＝b...
cosan-an=cosbn,得 an = cosan-cosbn ≤ 1-cosbn = 2[sin(bn/2)]^2 an/bn<an/|2sin(bn/2)| ≤ |sin(bn/2)| 则 ∑<n=1,∞>an/bn/ 收敛,
级数的比较判别法..如图,该题的第一问我搞不明白啊。答案里那个an<bn,我真的迷,不懂求解释。我自己能得出0<an<1,其它没办法了。mark。上面俩方法,贼秀…
【解析】(1)由a2=,a3=, 可得b2=o-,b=co-,公比为q= 由=b1·b3解得b1=1, 数列{bn}的通项公式为bn=(-2,-1 (2)证明:设存在递减的数列{an},使得{bn}是 无穷等比数列, 则0.a2a1,此时cosa2c0sa10, 公比q=cos2,1,cosan=cosa1·qn-1,考虑不等式 cosal cosa1·q-1,1, 当n1-logq(cosa...
已知等差数列{an}的公差为θ,bn=cosan,数列{b}的前n项和为Sn,S=\(S_n|n∈N^*\),若存在a1=2,使得S={a,b,c},则θ可能的取值为(
(bn-an)]/2]=(1/2)[(bn)^2-(an)^2]即an<(1/2)[(bn)^2-(an)^2]所以an/bn<(1/2)[bn-(an)^2/bn]因为an/bn<1，所以(an)^2/bn<an 所以∑(an)^2/bn收敛，且∑bn收敛。所以∑(1/2)[bn-(an)^2/bn]又因为an/bn<(1/2)[bn-(an)^2/bn]所以∑(an/bn)收敛 ...
BG Group to sell Comgøs holding to Cosan for $1.8bn.The article reports that BG Group and Cosan has signed a memorandum of understanding for the latter's acquisition of BG Group for about 1.8 billion U.S. dollars.EBSCO_bspEngineer