Sn=1+1/2+1/3+.+1/n,f(n)=S2n+1-Sn+1,求f(n)>log2(m-1)-log(m-1)2
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
赞 nickmemy 幼苗
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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
1年前
8
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