Proving a sequence (defined recursively) is monotonically decreasing

[Woolyabyss]

Homework Statement

A sequence (an) of real numbers is defined by a(1) = 6, a(n) = (1/4)(2*a(n−1) − 3) for n >= 2. where n is a natural number

show the sequence is monotonically decreasing

Homework Equations

None

The Attempt at a Solution

I tried to prove it by induction.
Let P(n) be the predicate that a(n) < a(n-1)[/B]

base case

p(2)...

a(1) = 6
a(2) = (1/4)*(2*6-3) = 9/4
a(2) < a(1) ... base case is true

induction step

assume p(k) to be true

a(k) < a(k-1)

now we try to prove p(k+1) is true using or assumption about p(k)

a(k+1) = (1/4)*(2(k) - 3)

rearranging

a(k) = 2*a(k+1) - 3

using our assumption

a(k-1) < 2*a(k+1) - 3

I'm not sure where to go from here I can't think of anyway of relating a(k) and a(k+1) with an inequality.
any help would be appreciated.

 

[pasmith]

Consider the recurrence $b_n = \frac12 b_{n-1}$ with $b_1 > 0$. Is that sequence monotonic decreasing?

Unfortunately your recurrence $a_n = \frac12 a_{n-1} - \frac34$ has an extra term on the right, but a linear change of variable to $b_n = a_n + C$ for some constant $C$ should remove it.

 

[Woolyabyss]

Consider the recurrence $b_n = \frac12 b_{n-1}$ with $b_1 > 0$. Is that sequence monotonic decreasing?

Unfortunately your recurrence $a_n = \frac12 a_{n-1} - \frac34$ has an extra term on the right, but a linear change of variable to $b_n = a_n + C$ for some constant $C$ should remove it.

so if we go back to a(k+1) = a(k)/2 - 3/4

let b(n) = a(k) - 3/2

so that a(k+1) = (1/2)*b(n)

b(n) is greater than a(k+1)

i.e.
a(k)/2 - 3/4 > a(k+1)

so I've got the an inequality in terms of the two terms of the sequence I need

rearranging

a(k) - 2a(k+1) > 3/2

i.e a(k) > 2a(k+1)

a(k+1) is positive

so a(k) > a(k+1)

would this be correct?

 

[haruspex]

so if we go back to a(k+1) = a(k)/2 - 3/4

let b(n) = a(k) - 3/2

so that a(k+1) = (1/2)*b(n)

I think you mean b(k) = a(k)-3/2, and the next line should be a recurrence relation on the b's, no a terms in sight.

 

[Woolyabyss]

I think you mean b(k) = a(k)-3/2, and the next line should be a recurrence relation on the b's, no a terms in sight.

I tried getting rid of the a's

b(k) = a(k)-3/2

b(k+1) = a(k+1) - 3/2

b(k+1) = (1/4)(2*a(k) -3) -3/2

b(k+1) = (1/2)*b(n) -3/2

I've got a recurrence relation in b now but I'm not sure where to go from here other than saying

2*b(k+1) = b(n) -3

b(n) - 2*b(k+1) = 3

 

[haruspex]

I tried getting rid of the a's

b(k) = a(k)-3/2

b(k+1) = a(k+1) - 3/2

b(k+1) = (1/4)(2*a(k) -3) -3/2

b(k+1) = (1/2)*b(n) -3/2

I've got a recurrence relation in b now but I'm not sure where to go from here other than saying

2*b(k+1) = b(n) -3

b(n) - 2*b(k+1) = 3

You've chosen the wrong C. Start again, substituting b_n-C for a_n everywhere in the recurrence relation for a_n (being careful not switch between k and n this time) and post the equation you get for the b sequence (with C still in it).

 

[HallsofIvy]

I don't see any good reason to introduce this new "b" sequence. To show that a sequence is increasing, it is enough to show that $a_{n+1}- a_n> 0$.