Bounded sequence $\{a_n\}_n$ such that $a_n < \frac{a_{n−1} + a_{n+1}}{2}$. Is $\{a_n\}_n$ convergent?

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.