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Let $\{a_n\}_n$ be a sequence of numbers in the interval $(0, 1)$ with the property that $$a_n < \frac{a_{n−1} + a_{n+1}}{2}$$ for all $n = 2, 3, 4,\dots$. Show that this sequence is convergent.

My attempt:

We can write the inequality as

$a_n - a_{n-1} < a_{n+1} - a_n$

So, sequence {$s_n$} = $a_{n+1} - a_n$ is monotonic and since -1<$s_n$<1 , it is also bounded and hence convergent.

Sequence {$a_n$} is bounded and by Bolzano-Weierstrass property has a convergent subsequence {$a_{n_k}$}.

Applying Cauchy sequence property on this convergent subsequence, we have for every $\epsilon$ >0, there is $N_0$, such that $|a_{n_l} - a_{n_k}|< \epsilon$ for all $l>k>N_0$

I feel like, from here I should have been able to prove this, but unfortunately I am stuck. Please help.

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  • $\begingroup$ I think you should be able to prove that $a_n$ is monotonic (decreasing). $\endgroup$ May 3, 2018 at 10:13

2 Answers 2

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If $(a_n)_n$ has a finite upper bound $M$ then $(a_n)_n$ is decreasing, i.e. $a_n\geq a_{n+1}$ for all $n\geq 1$. Otherwise $a_{n+1}>a_n$ for some $n$ and, for $k>1$, $$M\geq a_{n+k}=(a_{n+k}-a_{n+k-1})+(a_{n+k-1}-a_{n-k-2})+\dots+(a_{n+1}-a_n)+a_n\\> k\underbrace{(a_{n+1}-a_n)}_{>0}+a_n\to +\infty$$ as $k\to +\infty$. Contradiction!

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I'm not sure if $s_n \in (-1,1)$, but it's clear that $|s_n| \le |a_n| + |a_{n+1}| \le 1 + 1 = 2$, so you still have the convergence of $s_n$. Denote $s = \lim\limits_{n\to\infty} s_n$.

  • If $s > 0$, $s_N > 0$ for sufficiently large $N$, so $a_{n+1} = a_n + s_n > a_n$ for all $n \ge N$. $(a_n)$ is monotone and bounded, so the Monotone Convergence Theorem implies that it's convergent, but $s_n = a_{n+1}-a_n \to 0$ as $n \to \infty$, contradiction.
  • A similar argument shows that $s$ cannot be negative.
  • So $s = 0$. As $(s_n)$ is strictly increasing and converges to zero, $(s_n)$ is negative. This shows that $(a_n)$ is strictly decreasing, so the Monotone Convergence Theorem implies that it's convergent.
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  • $\begingroup$ Since $a_n$ ∈ (0,1) $s_n$ will lie in (-1,1). Thanks for your solution. $\endgroup$
    – Lord KK
    May 3, 2018 at 10:20
  • $\begingroup$ @AloknathFurr My bad. I misread $(0,1)$ as $(-1,1)$. Thanks for correcting me. $\endgroup$ May 3, 2018 at 10:22
  • $\begingroup$ @GNU Supporter: What guarantees that $s_n$ is strictly increasing in the third case? $\endgroup$ Jul 16, 2018 at 1:56
  • $\begingroup$ @Mike See OP's inequality $$a_n - a_{n-1} < a_{n+1} - a_n$$ and his definition of $s_n := a_{n+1} - a_n$. $\endgroup$ Jul 16, 2018 at 22:16

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