a=2

a(n+ 1)= -2/an +2 + 1=3？ -2/an=(3an-2)/an

a(n+ 1)- 1 =(3an-2-an)/an =(2an-2)/an = 2(an- 1)/an

a(n+ 1)-2 =(3an-2-2an)/an =(an-2)/an

[a(n+ 1)- 1]/[a(n+ 1)-2]= 2a(n- 1)/(an-2)

[a (n+1)-1]/[a (n+1)-2]/a (n-1)/(an-2) = 2, which is a fixed value.

(a 1- 1)/(a 1-2)=(3- 1)/(3-2)= 2

The sequence {(an- 1)/(an-2)} is a geometric series with 2 as the first term and 2 as the common ratio.

(an- 1)/(an-2)=2 2？ =2?

an=(2？ - 1)/(2? - 1)

When n= 1, a 1=(2? -1)/(2- 1)=3, which also satisfies the expression.

The general formula of the sequence {an} is an=(2? - 1)/(2? - 1)

dn=(2an-4)/(5an-7)

=[2(2? - 1)/(2? - 1) -4]/[5(2? - 1)/(2? - 1) -7]

=[2(2? - 1) -4(2? - 1)]/[5(2? - 1) -7(2? - 1)]

=2/(3 2? +2)

2? Constant > 0,2/(3 2? +2) constant > 0, dn>0

d 1 = 2/(3 ^ 2+2)=？

d(n+ 1)/dn=[2/(3 2？ -2)]/[2/(3 2? +2)]

=(3 2? -2)/(3 2? -2)

=? (3 2? -4)/(3 2? -2)

=? (3 2? -2-2)/(3 2? -2)

=? ([ 1- 2/((3 2? -2)]

=? - 1/(3 2? -2)

& lt？

Tn=d 1+d2+...+dn

& lt？ +? ? +...+?

=? ( 1-)/( 1-? )

=? ( 1-)

=? -

& lt？

& lt4/7

Inequality still exists.