Here is one such sequence...
$$5, 4, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...$$
the above is an infinitely long sequence of natural numbers such that $a_n \ge a_{n+1}$ for $n \ge 1$. It is sufficient to use this sequence as proof that such a sequence exists. A proof of this kind would be called a "constructive existence proof," and the above sequence would be a "witness" to the statement you are trying to prove.
If you were searching for an infinitely long sequence of natural numbers $a_1, a_2, a_3, ...$ such that $a_n > a_{n+1}$ for $n \ge 1$, then I would say no such sequence exists by the well ordering principle, which states that every nonempty subset of the natural numbers has a least element. In that event, you would be looking for a way to arrange the natural numbers such that they descend forever without ever encountering a least element beyond which there was no lesser number. Of course, the natural numbers have a least element, that is, the number $0$, so any "process" or function that generates a descending sequence of natural numbers will either terminate before it encounters $0$ or encounter $0$ and have to stop. In either case, the sequence is not infinite.
