Sn=1+1/2+1/3+.+1/n,f(n)=S2n+1-Sn+1,求f(n)>log2(m-1)-log(m-1)2
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/09/23 20:21:38
Sn=1+1/2+1/3+.+1/n,f(n)=S2n+1-Sn+1,求f(n)>log2(m-1)-log(m-1)2恒成立,m的取值范围
S(2n+1)=1+1/2+1/3+.+1/(2n+1)
S(n+1)=1+1/2+1/3+.+1/(n+1)
f(n)=1/(n+2)+1/(n+3)+……+1/(2n+1)
f(n+1)=1/(n+3)+……+1/(2n+1)+1/(2n+2)+1/(2n+3)
f(n+1)-f(n)=1/(2n+2)+1/(2n+3)-1/(n+2)
>1/(2n+2)+1/(2n+3)-2/(2n+4)>0
limf(n)=1/(n+2)+1/(n+3)+……+1/(2n+1)=ln(2n+1)-lnn=ln2
n->∞
m
S(n+1)=1+1/2+1/3+.+1/(n+1)
f(n)=1/(n+2)+1/(n+3)+……+1/(2n+1)
f(n+1)=1/(n+3)+……+1/(2n+1)+1/(2n+2)+1/(2n+3)
f(n+1)-f(n)=1/(2n+2)+1/(2n+3)-1/(n+2)
>1/(2n+2)+1/(2n+3)-2/(2n+4)>0
limf(n)=1/(n+2)+1/(n+3)+……+1/(2n+1)=ln(2n+1)-lnn=ln2
n->∞
m
已知数列{an}的前n项和为Sn=1+2+3+4+…+n,求f(n)= Sn /(n+32)Sn+1的最大值
已知等差数列{an}满足 a1+a(2n-1)=2n设Sn是数列{1/an}的前n项和,记f(n)=S2n-Sn(n属于
等差数列{an}a1=1前n项和为Sn且S2n/Sn=4n+2/n+1 (1)求an通项公试
等差数列an中,a1=1前n项和Sn,满足条件S2n/Sn=4n+2/n+1,求an通项
数列有一点没看懂bn=1/n,Sn=1+1/2+.+1/n所以,Tn=S2n-Sn=1/(n+1)+1/(n+2)+1/
在等差数列an中,a1=1,前N项和SN满足条件s2n/sn=4n+2/n+1,n=1,2,3.
等差数列{Sn}中,a1=1,前n项和Sn满足条件 S2n/Sn=4,n=1,2,.
等差数列{an},a1=1,前n项和Sn,S2n/Sn=4
等差数列,a1=1,前n项和满足S2n/Sn=(4n+2)/(n+1) n属于正整数 求an数列
已知等比数列an,m=(sn-r,2^n-1),N=(2.1)且向量m//n求常数r
已知数列{An}的前n项和为Sn,A2n=n+1(n∈N*),S2n-1=4n^2-2n+1(n∈N*),求数列{An}
等差数列(an),前n项和为Sn.(1)Sm=n,Sn=m.求Sm+n的值(2)Sm=Sn(m不等于n)求Sm+n的值