How can the convergence of a sequence be proven using the Squeeze Law method?

[ak123456]

1. The problem statement, all variablproves and given/known data
X0=1, Xn+1=1/4 (1+2/Xn) for n=0,1,2,3,4...
prove that this sequence converges ,and find the limit

Homework Equations

The Attempt at a Solution

I though that Squeeze law might be used ,and let L=1/4(1+2/L) L=0.8437
and the sequence is not monotonic
so , how can i prove it next ?

 

[Mark44]

I would calculate a few terms in the series to see if I could come up with a non-recursive formula for xn.

 

[ak123456]

I would calculate a few terms in the series to see if I could come up with a non-recursive formula for xn.

but how can i apply Squeeze law.

 

[Mark44]

First things first. Try to find a non-recursive formula for x_n, and then worry about the technique to use to see if the sequence converges or not.

 

[jbunniii]

[edit] oops, made a mistake, deleting it so I can fix it

 

[grado]

I'm interested how this is solved. Has anyone done it? If so, how.

 

[grado]

I'm interested how this is solved. Has anyone done it? If so, how.

I've just been having a pop at this and decided to turn it into two sequences, one increasing, one decreasing, where:

X_n+1 = 1/4 + 2/(1+2/X_n)
With X_0 = 1 or 3/4

Would it be sufficient to show that one or both of these are bounded and monotonic?

 

[jbunniii]

I think I finally found a solution, but it's ugly and hopefully someone knows a much cleaner way.

Let L be the positive solution to

L = \frac{1}{4}\left(1 + \frac{2}{L}\right)

or equivalently

4L^2 - L - 2 = 0

Thus, L \approx0.84307.

If (x_n) has a limit, then it must be L because it's easy to see that x_n is never negative.

The hard part is proving that (x_n) has a limit. I shall do this by proving that the sequence

y_n = x_n - L

has a limit of 0. To do this, I shall find a constant C < 1 such that

\left|\frac{y_{n+1}}{y_n}\right| < C

for all n.

First, I need a small lemma, namely:

|y_n| < 0.2 for all n

This will be proved inductively. Clearly it's true for n = 0, because

|y_0| = |x_0 - L| \approx |1 - 0.84307| \approx 0.15693

Now suppose that |y_n| < 0.2; we must show that this implies |y_{n+1}| < 0.2. Indeed, |y_n| < 0.2 implies that L - 0.2 < x_n < L + 0.2. Therefore,

\frac{2}{L+0.2} < \frac{2}{x_n} < \frac{2}{L-0.2}

and this means

\frac{1}{4}\left(1 + \frac{2}{L+0.2}\right) < \frac{1}{4}\left(1 + \frac{2}{x_n}\right) < \frac{1}{4}\left(1 + \frac{2}{L - 0.2}\right)

This is the same as

0.72935 < x_{n+1} < 1.0275

and we also have L - 0.2 < 0.72935 and 1.0275 < L + 0.2, so

L - 0.2 < x_{n+1} < L + 0.2

or equivalently

|y_{n+1}| < 0.2, concluding the proof of the lemma.

For the rest of the proof, we substitute x_n = y_n + L into the original recurrence relation, cross-multiply the denominator, and use the fact that 4L^2 - L - 2 = 0 to simplify the result to

4y_{n+1}y_n + 4L(y_n + y_{n+1}) = y_n

and solving for y_{n+1} we get

y_{n+1} = \frac{y_n(1 - 4L)}{4(y_n+L)}

Then, using the fact that

\frac{1}{|y_n+L|} = \frac{1}{|x_n|} = \frac{1}{x_n} < \frac{1}{L - 0.2}

we have

|y_{n+1}| = \frac{|y_n| \cdot |1 - 4L|}{4|y_n+L|} < \frac{|y_n| \cdot |1 - 4L|}{4|L - 0.2|} = |y_n| \cdot 0.92225

Thus

\frac{|y_{n+1}|}{|y_n|} < C

where C = 0.92225, concluding the proof.

It ain't pretty, but it gets the job done, unless I screwed up somewhere.

By the way, there is nothing magic about the number 0.2. At some point I concluded that I needed a number less than 0.25 but big enough to handle the variation of the sequence about its limit, and I chose 0.2 after "cheating" by plotting the sequence in Matlab.