Convergence when $2x_{n+2}+x_n\le 3x_{n+1}$ I have a sequence of real numbers $x_n$ which is bounded and $2x_{n+2}+x_n\le 3x_{n+1}, \forall n\in Z_+$ and I have to prove this sequence is convergent.
My initial idea was that if the condition implies $x_n$ is monotone, then from the Monotone Convergence Theorem, $x_n$ is convergence, but I don't think the condition is enough to prove this.
I also tried with $y_n=x_{n+1}-x_{n}$:
$2x_{n+2}+x_n\le 3x_{n+1} \implies 2(x_{n+2}-x_{n+1})\le x_{n+1}-x_n\implies 2y_{n+1}\le y_n$
but this is not enough to prove monotony, $y_n$ may be  negative.
I would appreciate any help.
 A: Hypothesis : Let us first assume that $x_1 \leq x_0$ (see Edit below)
Let us assume that :
$$(\text{property} (P_n)) : \ \ \ x_{n+1} \leq x_n.$$
The initial inequality can be written :
$$x_{n+2} \leq \underbrace{\dfrac32 x_{n+1}-\dfrac12 x_n}_{z_n}\tag{1}$$
$z_n$ is a weighted mean with a repulsive weight $-\dfrac12$ on $x_n$ and an attractive weight $\dfrac12$ on $x_{n+1}$. Therefore (geometricaly thinking) $z_n$ is outside line segment $[ x_{n+1},x_n]$ on the side of $x_{n+1}$ :
$$z_n\leq x_{n+1} \leq x_n.$$ 
Due to (1), we can deduce that
$$x_{n+2} \leq x_{n+1} \ \ \ \text{which is property } \ (P_{n+1})$$
As ($P_0$) is true, we have therefore proven by induction that $x_n$ is a decreasing sequence.
As it is bounded (here from below) it converges to its infimum.

Edit : Now, how can we consider the general case, i.e., do not need to have the hypothesis in front of this answer.
Very easily :either $x_n$ is always increasing and, as it is bounded, it will converge to its supremum, or there exists a value $n_0$ of $n$ such that :
$$x_{n_0+1} \leq x_{n_0}$$
But in this case, we are back to the initial situation and our proof by induction will begin at $n_0$ instead of $0$ ! ($x_n$ has been increasing in a first phase, then decreasing "for the rest of its life").
A: If all $y_n$ are non-negative then $(x_n)$ is increasing (and bounded above) and therefore convergent.
Otherwise $y_N < 0$ for some $N$, and then $2y_{n+1}\le y_n$ implies that $y_n < 0$ for all $n \ge N$. In that case $(x_n)_{n \ge N}$ is decreasing (and bounded below) and therefore also convergent.
A: A more appropriate choice might be $y_n=2x_{n+1}-x_n$. Then the condition gives $y_{n+1}\leq y_n$, so $y_n$ is convergent, say $y_n\to y$. Solving the recurrence:
$$x_n=\frac{1}{2^{n-2}}x_1+\frac{2^{n-1}y_{n-1}+2^{n-3}y_{n-3}+\ldots+y_1}{2^{n-1}}$$
and from Cesaro-Stolz:
$$\lim_{n\to \infty}x_n=\lim_{n\to \infty}\frac{2^{n-2}y_{n-1}+2^{n-3}y_{n-3}+\ldots+y_1}{2^{n-1}}=\lim_{n\to \infty} \frac{2^{n-1}y_n}{2^n-2^{n-1}}=y$$
A: Let $d_n = 2 x_{n+1}-x_n$, then $d_{n+1} \le d_n$ and since $x_k$ is bounded we have
$d_n \downarrow d^*$ for some $d^*$.
Since $x_{n+1} = {1 \over 2} (x_n + d_n)$, we can solve the recurrence to get 
$x_n = {1 \over 2^n} x_0 + {1 \over 2}\sum_{k=0}^n {1 \over 2^k} d_{n-k}$,
or more usefully,
$x_{n+m} = {1 \over 2^n} x_m + {1 \over 2}\sum_{k=0}^n {1 \over 2^k} d_{m+n-k} = {1 \over 2^n} x_m + {1 \over 2}\sum_{k=0}^n {1 \over 2^k} (d_{m+n-k}-d^*) +(1-{1 \over 2^{n+1}})d^*$.
Since $d_n \downarrow d^*$, it is not hard to see that for large $m$, and $n \ge 0$ the
quantity $\|x_{n+m}-d^*\|$ is arbitrarily small and
hence $x_n \to d^*$.
