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

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?

• You say you tried showing that it converged, but your attempt failed. Can you upload some screenshot of your attempt to solve the problem? This would help us understand where you went wrong. – Noble Mushtak Jan 1 at 20:08
• I honestly didn't get to anything worth uploading, I'm mostly stuck and trying different things. All I managed is to show is that it doesn't diverge to $+\infty$. I also tried splitting into cases for $a_0$ (positive or negative). – Belkan Jan 1 at 20:21
• Perhaps this would help math.stackexchange.com/questions/1842784/…. – Song Jan 1 at 20:45
• Here is a suggestion. As in the wrong answer below $a_n$ is bounded from above, say by a constant $a>0$. If $a_n$ does not go to $-\infty$ then there exists $M>0$ such that the interval $[-M,a]$ contains infinitely many terms of the sequence. But since $[-M,a]$ is compact it follows that $a_n$ has a convergent subsequent with limit say $l\in [-M,a]$. I am trying to figure out how to use this to prove that $l$ is the limit of $a_n$ – mouthetics Jan 1 at 21:00
• I tried going that way, but my problem was that $a_n\le\frac{1}{2}(a_{n-1}+a_{n-2})$ no longer meant the same thing (the indexes of the subsequent aren't necessarily sequential). However it seems like a good direction. – Belkan Jan 1 at 21:05

## 2 Answers

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 $$a_n\to -\infty.$$

• "If the sequence is not convergent to a real number $L$ then $a_n\to \infty$ or $a_n\to-\infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all? – Noble Mushtak Jan 1 at 20:16
• @NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly. – Jack D'Aurizio Jan 2 at 0:37

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.