Prove that $a_{n+2}=0.5(a_n+a_{n+1})$, $a_1=2, a_2=5$ has a limit, and find it I have the following sequence: $a_{n+2}=0.5(a_n+a_{n+1})$, $a_1=2, a_2=5$, and I need to prove that $\lim\limits_{n\to \infty}a_n$ exists, and find it.
I don't know ho to prove that the limit exists, since the sequence neither decreasing nor increasing.
In addition, I don't know how to find it, since the equation I am getting is $L=0.5(L+L)$, whitch is true for all $L$.
 A: $a_1 = 2$, $a_2 = 5$
$$2a_3 = a_1 + a_2$$
$$2a_4 = a_2 + a_3$$
$$2a_5 = a_3 + a_4$$
$$\cdots$$
$$2a_{n - 1} = a_{n- 2} + a_{n - 3}$$
$$2a_n = a_{n- 1} + a_{n - 2}$$
Add all equations, you will get
$$2a_n  + a_{n - 1} = a_1 + 2a_2$$
Since you can show that the sequence is Cauchy and hence converges as I explained in the comment above, $a_n \to L$ and $a_{n - 1}\to L$ as $n \to \infty$
Hence, $$2L + L = a_1 + 2a_2$$
or $$L = \dfrac{2 + 2(5)}{3} = 4$$
A: As mentioned in comments, the sequence may be shown to be Cauchy and hence convergent.
This is a second-order homogeneous linear recurrence. The characteristic equation is
$$2\lambda^2-\lambda-1=(2\lambda+1)(\lambda-1)=0$$
with roots $1$ and $-\frac12$. Therefore the general equation is $a_n=p+q(-1/2)^n$ where $p$ and $q$ are determined from the initial conditions; here $p=q=4$. Since $(-1/2)^n$ tends to zero as $n\to\infty$, the limit is $p=4$.
A: Here I leave a sketch. I may return and provide the details if someone asks for it when I am free.
Let $x_n = a_{2n+1}$ for each $n \ge 1$, and $y_n = a_{2n}$ for each $n \ge 1.$
Show that $(x_n)_{n \ge 1}$ and $(y_n)_{n \ge 1}$ are monotone and bounded and hence conclude that each of them converge to a finite limit. Let $l, l'$ be the limits of these two sequences respectively.
Now, use $a_{n+2} = \frac{a_n + a_{n+1}}{2}$ to show that $l = l'.$
Conclude from here that $(a_n)_{n \ge 1}$ converges.
Also you can find the value of the limits $l , l'$ and hence find the point where $(a_n)_{n \ge 1}$ converges.
