Correct answer: (A) The closure of the span of all fₙ is H.
Explanation
If a vector is orthogonal to every fₙ, its coefficients obey $a_{n+1}=(n+1)a_n$, hence $a_n=n!a_1$. Square summability forces a₁ = 0, so all coefficients vanish. The orthogonal complement of the span is therefore zero and the span is dense, proving A. For B and D, f = e₁ is a counterexample. Consecutive fₙ also have nonzero inner product, so C fails.
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