Question

已知数列an的通项公式an=log2 n+1/n+2(n属于正整数），设an的前n项和为sn，则使sn<-5c成立的自然数n为多少

应该是sn<-5

Answer 1

a1=log2 2/3 a2=log2 3/4.....

a1+a2+...an=log2 (2/3)(3/4)(4/5)...(n+1/n+2)

=log2 2/(n+2)=Sn

sn<-5 即 2/(n+2)<2^(-5)

算得 n>62

n=63,64......