已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1 (1)求证f(n+1)>f(n) (2)试确定m的取值已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1(1)求证f(n+1)>f(n)(2)试确定m的取值范
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![已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1 (1)求证f(n+1)>f(n) (2)试确定m的取值已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1(1)求证f(n+1)>f(n)(2)试确定m的取值范](/uploads/image/z/15250436-44-6.jpg?t=%E5%B7%B2%E7%9F%A5sn%3D1%2B1%2F2%2B1%2F3%2B%E2%80%A6%E2%80%A6%2B1%2Fn%EF%BC%88n%E2%88%88N%EF%BC%89%E4%B8%94f%EF%BC%88n%EF%BC%89%3Ds2n%2B1-sn%2B1+%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81f%EF%BC%88n%2B1%EF%BC%89%EF%BC%9Ef%EF%BC%88n%EF%BC%89+%EF%BC%882%EF%BC%89%E8%AF%95%E7%A1%AE%E5%AE%9Am%E7%9A%84%E5%8F%96%E5%80%BC%E5%B7%B2%E7%9F%A5sn%3D1%2B1%2F2%2B1%2F3%2B%E2%80%A6%E2%80%A6%2B1%2Fn%EF%BC%88n%E2%88%88N%EF%BC%89%E4%B8%94f%EF%BC%88n%EF%BC%89%3Ds2n%2B1-sn%2B1%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81f%EF%BC%88n%2B1%EF%BC%89%EF%BC%9Ef%EF%BC%88n%EF%BC%89%EF%BC%882%EF%BC%89%E8%AF%95%E7%A1%AE%E5%AE%9Am%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83)
已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1 (1)求证f(n+1)>f(n) (2)试确定m的取值已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1(1)求证f(n+1)>f(n)(2)试确定m的取值范
已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1 (1)求证f(n+1)>f(n) (2)试确定m的取值
已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1
(1)求证f(n+1)>f(n)
(2)试确定m的取值范围,使得对于一切大于1的自然数n,不等式f(n)>﹝logm(m+1)2-11/20(logm-1m)2﹞恒成立
已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1 (1)求证f(n+1)>f(n) (2)试确定m的取值已知sn=1+1/2+1/3+……+1/n(n∈N)且f(n)=s2n+1-sn+1(1)求证f(n+1)>f(n)(2)试确定m的取值范
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
2.第二问我看不懂这个.底数和真数.
不过我额可以告诉卤煮思路.
limf(n)=1/(n+2)+1/(n+3)+……+1/(2n+1)=ln(2n+1)-lnn=ln2
n->∞
所以这个函数的极限是IN2.知道左边了.右边你解不等式就可以了.只有一个未知数M.不好意思了.我看不清真数和底数.