Showing a sequence is less than another sequence (sequences and series question)

[cloud360]

!Showing a sequence is less than another sequence (sequences and series question)

Homework Statement

[PLAIN]http://img263.imageshack.us/img263/385/sandq1.gif

Homework Equations

The Attempt at a Solution

i re arranged the equation and wrote in terms of a_n

to get a_n=-4a_n+1/a_n+1-7

and a_n+1 can not be 7

but i don't know how to show that, whatever value of a_n+1 i sub, it will always be less than a_n

 

[cloud360]

is this proof?

Let a_n = 3 + x where x is positive.

a_n+1= 7(3 + x) / (3 + x + 4) = (21 + 7x)/(7 + x)

= (21 + 3x + 4x) / (7 + x) = 3 + 4x/(7 + x) > 3

re arrange and factorise ==> a_n>0 and a_n>3

This is because the last part 4x/(7 + x) must be positive if x is positive.

Now suppose a_n+1 = 7*a_n / (a_n + 4) <= a_n. What would follow?a_n > 3 ----> (a_n)² > 3*a_n ----> (a_n)^2 + 4*a_n > 7*a_n

Divide by (a_n + 4) to get a_n > 7a_n / (a_n + 4)

a_n > a_n+1 or a_n+1 < a_n