Prove that the sequence $\{y_n\}$ where $y_{n+2}=\frac{y_{n+1} +2 y_{n}}{3}$ $n\geq 1$, $0
Prove that the sequence $\{y_n\}$ where $y_{n+2}=\frac{y_{n+1} +2 y_{n}}{3}$ $n\geq 1$, $0<y_1<y_2$, is convergent by using subsequencial criteria, by showing $\{y_{2n}\}$ and $\{y_{2n-1}\}$ converges to the same limit. Find the limit also.
I can solve it by Cauchy sequence as $|y_m-y_n|\leq |y_{n}-y_{n+1}|+|y_{n+1}-y_{n+2}|\cdots +|y_{m-1}-y_m|\cdots$,
but here, we have to check convergence by using subsequencial criteria, by showing $\{y_{2n}\}$ and $\{y_{2n-1}\}$ converges to the same limit. Please help.  
 A: You could prove that the sequence is monotonous and bounded, precisely, using induction you can prove that the sequence is monotonously increasing and is bounded from above by $y_2$. This is sufficient to prove the convergence.
A: (1) To eliminate the trivial case, first, note that if any two terms are equal $y_{n+2}=y_{n+1}$ then the recurrence gives $y_{n+1}=y_n$ and, repeating the same argument, in the end $y_2=y_1$ therefore a constant sequence. But the premise $y_1<y_2$ excludes that case, so no consecutive terms are equal.
(2) $\min(y_{n+1}, y_n) \le y_{n+2}=\frac{y_{n+1} +2 y_{n}}{3} \le \max(y_{n+1}, y_n)$ implies that each term is in between the previous two, so in the end all terms are within $[y_1,y_2]$ therefore the sequence is bounded.
(3) Writing the recurrence as $y_{n+2}-y_{n+1}=\frac{-2}{3}(y_{n+1}-y_n)\,$, with both sides being non-0 per (1) implies that:
$$\operatorname{sgn}(y_{n+2}-y_{n+1}) = - \operatorname{sgn}(y_{n+1}-y_{n}) = \cdots = (-1)^n \operatorname{sgn}(y_2-y_1) = (-1)^n$$
(4) Writing the recurrence now as $y_{n+2}-y_n=\frac{1}{3}(y_{n+1}-y_n)\,$ implies that:
$$\operatorname{sgn}(y_{n+2}-y_n) = \operatorname{sgn}(y_{n+1}-y_{n}) = (-1)^{n+1}$$
For $n=2k$ this translates to $\operatorname{sgn}(y_{2k+2}-y_{2k}) = (-1)^{2k+1} = -1\,$, so $y_{2k+2} \lt y_{2k}$ thus the sequence $(y_{2k})$ is decreasing. Similarly, the subsequence $(y_{2k+1})$ is increasing. Since both subsequences are bounded per (2), both must converge.
(5) From (3) $y_{n+2}-y_{n+1}=\frac{-2}{3}(y_{n+1}-y_n)\,$ it follows that $\lim_{n \to \infty} (y_{n+1}-y_n) = 0$ therefore the odd and even subsequences must converge to the same limit.
(6) To find the actual limit, the recurrence can be written as $3y_{k+2}=y_{k+1}+2y_k$, then summing up for $k=1\,...\,n$ and telescoping leaves in the end $3y_{n+2}+2y_{n+1}=3y_2+2y_1$. Since it's been proved already that the limit $\lim_{n\to\infty} y_n = L$ exists, passing to the limit gives $5 L = 3 y_2 + 2 y_1\,$.
