数列an+1=an2+an(n+1为脚标),a1=1,求证1/(a1+1)+1/(a2+1)+1/(a3+1)+...+1(an+1)
人气:350 ℃ 时间:2020-07-23 00:08:57
解答
题目出错了,不等号方向反了,而且没有规定n的取值.
证:
假设当n=k(k∈N,且k≥1)时,ak>0,则当n=k+1时,a(k+1)=ak²+ak=ak(ak+1)
ak>0ak +1>0a(k+1)>0
k为任意正整数,因此an>0,即数列各项均>0
a(n+1)=an²+an
a(n+1)/an=an+1>1
a(n+1)>an,数列为递增数列.
a2=a1²+a1=1+1=2
n≥2时,an≥21/an≤1/2
a(n+1)=an²+an
1/a(n+1)=1/[an²+an]=1/[an(an +1)]=1/an -1/(an +1)
1/(an +1)=1/an -1/a(n+1)
1/(a1+1)+1/(a2+1)+...+1/(an +1)
=1/a1-1/a2+1/a2-1/a3+...+1/an -1/a(n+1)
=1/a1 -1/a(n+1)
=1 -1/a(n+1)
n≥1,n+1≥2a(n+1)≥20
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