It's supposed to be "The sequence is eventually periodic with period three", that is, for sufficiently large $n$, we have $a_n=a_{n+3}$

Maybe we can prove the problem with following steps. Step 1: The sequence has only finitely many terms $a_k$, such that $a_k \geq 1000$. (Just notice that sum of digits of a number is quite small in comparision with the number itself.) Step 2: The sequence modulo $9$ will sure to be like this: $2, 5, 8, 2, 5, 8, \cdots $, it implies that the sequence modulo $9$ will finally periodic with period three. The problem is done with the two steps above. Notice that for $m \leq 999$, its sum of digits can only be $2,11,20,5,14,23,8,17,26$ if $m \equiv 2,5,8 [9]$. With some analysis, we get the sequence should be $65,122,26,65,122,26, \cdots$ $\blacksquare$