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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.

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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.

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