Prove that $ x_n $ converges if $ x_{n+3}\leq\frac{x_{n+2}+3x_{n+1}+x_n}{5} $.

$ \left\{x_n\right\} $ is a sequence and $ x_{n+3}\leq\frac{1}{5}(x_{n+2}+3x_{n+1}+x_n) $ for all $ n\geq 1 $. Prove that $ \left\{x_n\right\} $ is convergent.

I have proved that $ x_n $ is bounded by using induction. Moreover, by simple calculation, $ 5x_{n+2}+4x_{n+1}+x_n $ is decreasing and has limits. I do not know how to go on, can you give me some hints or references?

Solution 1:

The sequence $(x_n)$ either diverges to $-\infty$, or it converges to a finite value.

(The proof works similarly as in $x_{n+m}\le \frac{x_n+x_{n+1}+\cdots+x_{n+m-1}}{m}$. Prove that this sequence has a limit.. We don't assume non-negativity here, which is why there are two possible cases.)

First define $$ y_n = \max(x_n, x_{n+1}, x_{n+2}) $$ and show that $(y_n)$ is a decreasing sequence.

Case 1: $(y_n)$ is not bounded below: Then $y_n \to -\infty$, which implies that $x_n \to -\infty$ since $x_n \le y_n$.

Case 2: $(y_n)$ is bounded below, and therefore convergent. Let $L = \lim_{n \to \infty} y_n$. We show that $(x_n)$ converges to the same value $L$.

Let $\epsilon > 0$. Then $$ x_n \le y_n < L + \epsilon $$ for all sufficiently large $n \ge N$. It follows that for $n \ge N+2$ $$ x_{n+1} \le \frac 15 (x_{n-2} + 3 x_{n-1} + \color{red}{x_{n}}) < \frac 15 (4(L + \epsilon) + x_n) \\ x_{n+2} \le \frac 15 (x_{n-1} + 2 x_{n} +\color{red}{x_{n}}+ x_{n+1}) < \frac 15 (4(L + \epsilon) + x_n) \\ x_{n+3} \le \frac 15 (\color{red}{x_{n}} + 3 x_{n+1} + x_{n+2}) < \frac 15 (4(L + \epsilon) + x_n) \\ $$ and therefore $$ y_{n+1} < \frac 15 (4(L + \epsilon) + x_n) \, . $$ But $(y_n)$ decreases to $L$, so that $$ L < \frac 15 (4(L + \epsilon) + x_n) \implies L - 4 \epsilon < x_n $$ for $n \ge N+2$.

We have shown that for all $\epsilon > 0$ there is an $N$ such that for $n \ge N+2$ $$ L - 4 \epsilon < x_n \le L + \epsilon $$ and that proves $\lim_{n \to \infty} x_n = L$.