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.

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

2 Answers 2


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!


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.
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.