# quick sequence question

#### slevvio

I was wondering if i could get some help with this question:

Define a sequence $$\displaystyle \{x_n\}_{n=1}^{\infty}$$ by $$\displaystyle x_1 = 1, x_2 = \frac{1}{2}$$ and$$\displaystyle x_n = \frac{2x_{n-1} + x_{n-2}}{4}$$for $$\displaystyle n \ge 3.$$ Use the Monotone Convergence Theorem to show that $$\displaystyle \{x_n\}_{n=1}^{\infty}$$ converges and find its limit.

Ok it is not too hard to prove by induction that $$\displaystyle 0 \le x_n \le 1$$. But I am having trouble proving the sequence is decreasing. I have

$$\displaystyle x_{n+1} - x_{n} = \frac{2x_n + x_{n-1}}{4} - x_n = \frac {x_{n-1} - 2x_n}{4}$$, but I have no guarantee that this is less than or equal to zero!

Any help would be appreciated!

#### kompik

I was wondering if i could get some help with this question:

Define a sequence $$\displaystyle \{x_n\}_{n=1}^{\infty}$$ by $$\displaystyle x_1 = 1, x_2 = \frac{1}{2}$$ and$$\displaystyle x_n = \frac{2x_{n-1} + x_{n-2}}{4}$$for $$\displaystyle n \ge 3.$$ Use the Monotone Convergence Theorem to show that $$\displaystyle \{x_n\}_{n=1}^{\infty}$$ converges and find its limit.

Ok it is not too hard to prove by induction that $$\displaystyle 0 \le x_n \le 1$$. But I am having trouble proving the sequence is decreasing. I have

$$\displaystyle x_{n+1} - x_{n} = \frac{2x_n + x_{n-1}}{4} - x_n = \frac {x_{n-1} - 2x_n}{4}$$, but I have no guarantee that this is less than or equal to zero!

Any help would be appreciated!
$$\displaystyle \frac{x_n}2 \le x_{n+1} \le x_n$$
by induction?

• slevvio

#### slevvio

thank you, i will try this! #### chisigma

MHF Hall of Honor
Let's write the difference equation as...

$$\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{2} =0$$ (1)

... so that its a constant coefficients homogeneous linear difference equation, and its solution is on the form...

$$\displaystyle x_{k} = c_{1}\cdot u_{k} + c_{2}\cdot v_{k}$$ (2)

The correponding characteristic equation is...

$$\displaystyle \xi^{2} - \frac{\xi}{2} - \frac{1}{2} =0$$ (3)

... the solutions of which are $$\displaystyle \xi=1$$ and $$\displaystyle \xi= -\frac{1}{2}$$, so that is...

$$\displaystyle u_{k} = 1^{k} =1$$

$$\displaystyle v_{k} = (-\frac{1}{2})^{k}$$ (4)

The 'initial condition' $$\displaystyle x_{0}=1$$ and $$\displaystyle x_{1} = \frac{1}{2}$$ give us $$\displaystyle c_{1}=\frac{2}{3}$$ and $$\displaystyle c_{2}=\frac{1}{3}$$ so that the solution of (1) is...

$$\displaystyle x_{k} = \frac{2}{3} + \frac{1}{3}\cdot (-\frac{1}{2})^{k}$$ (5)

... and is also $$\displaystyle \lim_{k \rightarrow \infty} x_{k} = \frac{2}{3}$$ ...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

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#### kompik

Let's write the difference equation as...

$$\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{2} =0$$ (1)
I think that (1) should be:
$$\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0$$,
i.e. 4 instead of 2 in the denominator.

#### slevvio

shouldnt the third term in your equation be over 4? cos the limit is definitely zero

#### slevvio

discovered this is a simple way to show this. Just subsitute the values for n and n+1 into one recurrence relation and subtract

• kompik

#### chisigma

MHF Hall of Honor
I think that (1) should be:
$$\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0$$,
i.e. 4 instead of 2 in the denominator.

The difference equation is then...

$$\displaystyle x_{k} - \frac{x_{k-1}}{2} - \frac{x_{k-2}}{4} =0$$ (1)

... and the characteristic equation...

$$\displaystyle \xi^{2} - \frac{\xi}{2} - \frac{1}{4}$$ (2)

... the solution of which are $$\displaystyle \xi= \frac{1 \pm \sqrt{5}}{4}$$, so that the solution of (1) is...

$$\displaystyle x_{k} = c_{1}\cdot (\frac{1+\sqrt{5}}{4})^{k} + c_{2}\cdot (\frac{1-\sqrt{5}}{4})^{k}$$ (3)

At this point it doesn't matter which are the 'initial conditions' because in any case is...

$$\displaystyle \lim_{k \rightarrow \infty} x_{k} = 0$$ (4)

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$