a(n+1)=f(1/an)=(2/an+3)/(3/an)=2/3+an所以{an}是以1为首项,2/3为公差的等差数列.an=(2n+1)/3Sn=(n2+2n)/31/3Sn=1/(n2+2n)=1/n(n+2)=1/2*(1/n-1/(n+2))
Tn=1/2*(1-1/(n+2))
a(n+1)=f(1/an)=(2/an+3)/(3/an)=2/3+an所以{an}是以1为首项,2/3为公差的等差数列.an=(2n+1)/3Sn=(n2+2n)/31/3Sn=1/(n2+2n)=1/n(n+2)=1/2*(1/n-1/(n+2))
Tn=1/2*(1-1/(n+2))