如果两个级数an^2和bn^2都收敛,试证明级数anbn绝对收敛-搜答案-神马作文网

an+an+1=2*bnbn+bn+1=2*an+12*bn+2*bn+1=4*an+1an+an+1+an+1+an+2=4*an+1an+an+2=2*an+1等差an=2n-1bn=2n

An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn

用比较判别法证明.经济数学团队帮你解答.请及时评价.

(an+bn)^2

算术几何均值不等式：|an|/n

如：an=n²,发散的,an＋bn=1/n,是收敛的,此时bn=－n²＋(1/n)还是发散的.

1.S2n+1=(A1+A2n+1)*(2n+1)/2=(2n+1)*An（由等差中项推导出来）,同理T2n+1=(2n+1)*Bn.所以An/Bn=S2n+1/T2n+1=（4n+4）/（2n+3）

S(2n-1)=(A1+A(2n-1))×(2n-1)/2=(A1+A1+(2n-2)d)×(2n-1)/2=(A1+(n-1)d)×(2n-1)=An×(2n-1)同理T(2n-1)=Bn×(2n-

绝对值an*bn

如图

n=√an*a(n+1)b(n+1)=√a(n+1)a(n+2)[b(n+1)/bn]^2=[a(n+1)*a(n+2)]/[a(n+1)*an]=a(n+2)/ana(n+2)=q^2*an

A11/B11=（33-3）/（22+3）=6/5A11=11/2(a1+a11)=11/2(2a1+10d)=11/2(2a6)=11a6A11=a1*11+11(11-1)d/2=11/2(2a1

令an-2n+11>0,解得：n

由已知得bn=[an+a(n+1)]/2a(n+1)²=bn×b(n+1)=[an+a(n+1)][a(n+1)+a(n+2)]/4[an+a(n+1)][a(n+1)+a(n+2)]=4a

Sn=3n-2,B1=S1=1Bn=Sn-S(n-1)=3(n>1)根据Bn=3的n次方乘An,当n=1时,A1=1/3当n>1时,An=3的n-1次幂再问：为什么Bn=Sn-S(n-1)=3再答：S

a(n)=aq^(n-1),a>0,q>0.a+aq=a(1)+a(2)=2[1/a(1)+1/a(2)]=2[1/a+1/(aq)]=2(q+1)/(aq),a=2/(aq),q=2/a^2,a(n

A9＝9a5(A9=a1+a2+...a9)(因为a1+a9=2a5,a2+a8=2a5,a3+a7=2a5,a4+a6=2a5)同样的道理,B9＝9b5所以a5/b5＝A9/B9＝28/21=4/3

由于有0

等差数列数列的性质a1+a[2n-1]=2an因为S[2n-1]=[(2n-1)(a1+a[2n-1])]/2=(2n-1)anT[2n-1]=[(2n-1)(b1+b[2n-1])]/2=(2n-1

s1=b1=3*a1=1,a1=1/3s2=s1+b2=1+9*a2=4,a2=1/3s3=s2+b3=4+27*a3=7,a3=1/9s4=s3+b4=7+81*a4=10,a4=1/27……﹛an