Calculus: Converge of a recursive series?

Question

I'm going crazy to solve this problem: I've a sequence defined by: $$x_1 = 1$$ $$x_2 = 2$$ $$x_{n+2} = \frac{1}{2}(x_{n}+x_{n+1})$$

And I have to prove that this sequence converges and what is its limit (I know what the limit is, but I don't know how to prove it...).

I tried with induction but I stuck during the induction step $n=n+1 \implies 1\leq \frac{1}{2}(x_{n+1}+x_{n+2})\leq 2 $

Tips?

Answer 1

Hint: Use induction to prove the following explicit formula:

$x_n = \dfrac{1}{3} \left(5 + 4(-\frac{1}{2})^n\right)$

And use that to show that the sequence converges to $\frac{5}{3}$.

Induction: Base case: $n=1, n=2$ give the proper formulae.

$2x_{n+2} = \frac{1}{3} \left(5 + 4(-\frac{1}{2})^{n+1} + 5 + 4(- \frac{1}{2})^n \right)$

$2x_{n+2} = \frac{1}{3} \left(5 - 2(-\frac{1}{2})^{n} + 5 + 4(- \frac{1}{2})^n \right)$

$2x_{n+2} = \frac{1}{3} \left(10 + 2 (- \frac{1}{2})^n\right)$

$x_{n+2} = \frac{1}{3} \left(5 + 4(- \frac{1}{2})^{n+2} \right)$.

Answer 2

Assume $x_i$ has the form of $a^i$, so we have:

$$x_{n+2} = \dfrac{1}{2}(x_{n+1}+x_n) \implies a^{n+2}=\dfrac{1}{2}(a^{n+1}+a^{n})$$

Assuming $a\neq 0$ we have

$$a^{2}-\dfrac{a}{2}- \dfrac{1}{2}=0$$

solving the equation

$$a = \dfrac{\dfrac{1}{2} \pm \sqrt{\dfrac{1}{4}+2}}{2} \implies a_{r1} =1 \land a_{r2} = -\dfrac{1}{2} $$

Now, we suppose that the general solution is a linear combination between this two answers

$$x_n=p a_{r1}^n + q a_{r2}^n \implies x_n = p + q \left(-\dfrac{1}{2}\right)^n$$

to get $p$ and $q$ simply replace the values with $n \in (1,2)$

$$x_1=p+q \left(-\dfrac{1}{2}\right) = 1$$ $$x_2=p+q \left(-\dfrac{1}{2}\right)^2 = 2$$

Solving the system we have $p=\dfrac{5}{3} \land q = \dfrac{4}{3} \therefore$

$$x_n = \dfrac{5}{3} + \dfrac{4}{3} \left(-\dfrac{1}{2}\right)^n$$

$$\lim_{n \to \infty}x_n = \lim_{n \to \infty}\dfrac{5}{3} + \dfrac{4}{3} \left(-\dfrac{1}{2}\right)^n = \dfrac{5}{3}$$

Answer 3

First, note that if $n \mapsto x_n$ satisfies the equation, then so does $n \mapsto x_n-c$ for any constant $c$.

Second, notice that if we start with $x_2=-{1 \over 2} x_1$, then we have $x_{n+1}=-{1 \over 2} x_n$ for all $n$, and so $x_n = -{1 \over 2}^{n-1} x_1$.

So, choose $c$ such that $(x_2-c) = -{1 \over 2} (x_1-c)$, which gives $c={5 \over 3}$.

Then we have $(x_n-{5 \over 3}) = (-{1 \over 2})^{n}(1-{5 \over 3}) $, or $x_n = {5 \over 3} + (-{1 \over 2})^{n-1}(1-{5 \over 3})$.

Note: $1, -{1 \over 2}$ are roots of the equation $x^2={1 \over 2}(x+1)$, and $x_1={5 \over 3}, x_2={5 \over 3}$ and $x_1 = 1-{5 \over 3}, x_2 = 2-{5 \over 3}$ are corresponding eigenvectors.