已知Sn是数列an的前n项和,且Sn=n的平方-4n+4

n=1时,a1=S1=1²-4×1+4=1
n≥2时,
Sn=n²-4n+4 S(n-1)=(n-1)²-4(n-1)+4
an=Sn-S(n-1)=n²-4n+4-(n-1)²+4(n-1)+4=2n-5
n=1时,a1=2-5=-3≠1
数列{an}的通项公式为
an=1 n=1
2n-5 n≥2
n=1时,1/a1=1/1=1 T1=1
n≥2时,1/an=1/(2n-5)
Tn=1/a1+1/a2+...+1/an= 1+1/(2×2-5)+1/(2×3-5)+...+1/(2n-5)
n=1时,
T(2n+1)-Tn=T3-T1
=1/a1+1/a2+1/a3-1/a1
=1/a2+1/a3
=1/(2×2-5)+1/(2×3-5)
=-1+1=0
T3-T1≤m/15 m/15≥0,m≥0
n≥2时,
T(2n+1)-Tn
=1+1/(2×2-5)+...+1/(2n-5)+1/[2(n+1)-5]+...+1/[2(2n+1)-5]-[1+1/(2×2-5)+...+1/(2n-5)]
=1/[2(n+1)-5]+1/[2(n+2)-5]+...+1/[2(2n+1)-5]
T[2(n+1)+1]-T(n+1)
=1+1/(2×2-5)+...+1/(2n-5)+1/[2(n+1)-5]+...+1/[2(2n+3)-5]-[1+1/(2×2-5)+...+1/(2n-5)+1/(2n-3)]
=1/[2(n+2)-5]+1/[2(n+3)-5]+...+1/[2(2n+3)-5]
[T[2(n+1)+1]-T(n+1)]-[T(2n+1)-Tn]
=1/[2(2n+3)-5]-1/[2(n+1)-5]
=1/(4n+1)-1/(2n-3)
=(2n-3-4n-1)/[(4n+1)(2n-3)]
=-2(n+2)/[(4n+1)(2n-3)]