# Show that sequence $(a_n)$ is convergent. Given that $a_{n}$ is bounded, $a_{n} \geq \frac{1}{2}({a_{n-1}}+{a_{n+1}})$

My attempt: I need to show that $$a_{n}$$ is monotonic.

$$a_{n+1}$$-$$a_{n}$$ $$\leq$$ $$a_{n}$$ - $$a_{n-1}$$.

If $${a_{n}}$$ is monotonically decreasing then $$a_{n}$$ $$\leq$$ $$a_{n-1}$$. This implies $${a_{n+1}}$$ $$\leq$$ $$a_{n}$$. Similarly for increasing.

## 1 Answer

You are on the right track. Note that the condition on the sequence essentially says that the sequence is concave; that is, the difference between adjacent terms (the slope of the plotted sequence) is non-increasing.

The inequality $$a_n\ge\tfrac12(a_{n-1}+a_{n+1})\tag1$$ is equivalent to $$a_{n+1}-a_n\le a_n-a_{n-1}\tag2$$ If for any $$n_0$$, $$a_{n_0}-a_{n_0-1}\le\lambda\lt0$$, then $$(2)$$ implies that for all $$n\ge n_0$$, $$a_n-a_{n-1}\le\lambda$$, which would contradict the boundedness of $$(a_n)$$. Thus, we must have $$a_n-a_{n-1}\ge0\tag3$$ That is, $$(a_n)$$ is non-decreasing and bounded above, hence, convergent.