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The sequence $(a_n)$ is bounded for $n=1, 2, \dots$, such that

$$a_n \leq \frac{1}{2} \left(a_{n-1} + a_{n+1}\right)$$

for $n \geq 2$. I want to prove the sequence $(a_n)$ converges.

Since I am told the sequence is bounded, I was trying to prove it is monotonic so that I could use a known theorem to claim it converges. However, I can prove it is bounded, but I am unable to prove it is monotonic. I tried to use series and use the fact that it is telescopic but nothing useful came out. Any ideas? Am I going in the wrong track?

Jimmy R.
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user186918
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  • Are the terms positive? Also, it's not necessarily monotonic. Actually, if you let each term be the largest it can be, the sequence isn't monotonic: e.g. $16,32,24,28,26,27,\ldots$. – Arthur Oct 24 '14 at 23:07

2 Answers2

2

Let's define the sequence $b_n = a_{n + 1} - a_n$. The condition $a_n \le \frac{1}{2}(a_{n - 1} + a_{n + 1})$ can be rearranged to $a_n - a_{n - 1} \le a_{n + 1} - a_n$, or put another way $b_{n - 1} \le b_n$. So the sequence $b_n$ is monotonically increasing. This implies that $sign(b_n)$ is eventually constant (either - or $0$ or +). This in turn implies that the sequence $a_{n + 1} - a_1 = b_1 + ... + b_n$ is eventually monotonic. More precisely, it's eventually decreasing if $sign(b_n)$ is eventually -, it's eventually constant if $sign(b_n)$ is eventually $0$, it's eventually increasing if $sign(b_n)$ is eventually +. Since the sequence $a_{n + 1} - a_1$ is also bounded, we get that it converges. This immediately implies that the sequence $a_n$ converges.

jflipp
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1

Let us fix $k\ge 1$.

Define $$ b_1 = a_k\\ b_2 = a_{k+1}\\ b_{n+1} = 2b_n - b_{n-1} $$

Then you can prove that $a_{n+k-1} \ge b_n$ and that $b_n = (a_{k+1}-a_k) n + 2a_k - a_{k+1}$.

As $a$ is bounded, $b_n\nrightarrow+\infty$ and $a_{k}\ge a_{k+1}$.

Hence the sequence is monotonic.

mookid
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