Question

an=4n-2,设bn=2/｛(2n+1)an｝,求bn的前n项和Tn,并证明Tn≥1/3

Answer 1

n=2/{2(2n+1)(2n-1)}=1/(2n-1)(2n+1)=(1/2)[1/(2n-1)]-[1/(2n+1)]

所以Tn=b1+b2+...+bn=(1/2)[1-1/3+1/3-1/5+...+1/(2n-1) -1/(2n+1)]=(1/2)[1-1/(2n+1)]=n/(2n+1)

又1/(2n+1)≤1/3,所以(1/2)[1-1/(2n+1)]≥1/3,即Tn≥1/3