Prove this sequence converges or diverges to $-\infty$

Question

Let $a_n$ be a sequence such that for every $n$: $a_n\le\frac{1}{2}(a_{n-1}+a_{n-2})$.

Prove that $a_n$ either converges to a real number $L$ or diverges to $-\infty$ $(L\in[-\infty,\infty))$.

I tried assuming it didn't diverge to $-\infty$ in order to show that in that case it must converge to a real number. I tried showing so by cauchy's convergence definition but I failed.

Any ideas?

Answer 1

So we have that $$a_3\leq \frac{1}{2}(a_{2}+a_{1})$$ $$a_4\leq \frac{1}{2}(a_{3}+a_{2})\leq 3a_2/4+a_1/4$$ $$a_5\leq \frac{1}{2}(a_4+a_3)\leq 7a_2/8+5a_1/8$$ $$\cdots$$ $$a_{n}\leq \frac{2n-5}{2^{n-2}}a_2+\frac{2n-7}{2^{n-2}}a_1.$$ If the sequence is not convergent to a real number $L$ then $a_n\to \infty$ or $a_n\to-\infty.$ However $$a_n\leq \frac{4n-12}{2^{n-2}}\max\{a_1,a_2\}$$ and so $a_n\to -\infty.$

Answer 2

Let $\delta_n = a_n-a_{n+1}$. From

$$ a_{n+1} \leq \frac{1}{2}a_n + \frac{1}{2}a_{n-1} $$ we get $$ -\delta_n = a_{n+1}-a_n \leq \frac{1}{2}\delta_{n-1} $$ so our sequence is free to decrease as fast (or as slow) as it likes, but if $a_{n}>a_{n-1}$ (i.e. $\delta_{n-1}<0$) then $a_{n+1}$ has to lie on the left of the midpoint of $[a_{n-1};a_n]$ and the subsequence of the terms which are greater than the previous one, say $\{a_{n_k}\}_{k\geq 1}$, is decreasing. We are temporary assuming this actually is a subsequence, i.e. that there are infinite $n$s such that $a_{n+1}>a_n$. Now we may start a dichotomy. If $a_{n_k}$ converges to $-\infty$, so does the original sequence. If $a_{n_k}\to L$, for any $k$ large enough we have $a_{n_k}\in(L-\varepsilon,L+\varepsilon)$ and $$ a_{n_k}\geq a_{n_k+1} \geq a_{n_k+2} \geq \ldots \geq a_{n_{k+1}-2} \geq \frac{a_{n_{k+1}-1}+a_{n_{k+1}-2}}{2}\geq\max(a_{n_{k+1}},a_{n_{k+1}-1}) $$ so for any $m\in[n_{k},n_{k+1}]$ we have $a_m\in(L-3\varepsilon,L+3\varepsilon)$ and $\{a_n\}_{n\geq 1}$ and $\{a_{n_k}\}_{k\geq 1}$ share the same limit. At last we have to deal with the case in which $\delta_n<0$ holds for a finite number of $n$, i.e. $\{a_n\}_{n\geq 1}$ is a weakly decreasing sequence from some point on. But again, a weakly decreasing sequence may only diverge to $-\infty$ or converge to a finite limit.