Q56·Section II (Mathematics)·56 of 190Stirling's growth versus a root that tames a logarithmLet an=(n!)1/(2n)a_n=(n!)^{1/(2n)}an=(n!)1/(2n) and bn=(1/n2)1/(3n)b_n=(1/n^2)^{1/(3n)}bn=(1/n2)1/(3n). Then the sequences: OptionsA(an)n,(bn)n∈ℓ∞(a_n)_n,(b_n)_n\in\ell^\infty(an)n,(bn)n∈ℓ∞B(an)n∈ℓ∞(a_n)_n\in\ell^\infty(an)n∈ℓ∞ and (bn)n∉ℓ∞(b_n)_n\notin\ell^\infty(bn)n∈/ℓ∞C(an)n∉ℓ∞(a_n)_n\notin\ell^\infty(an)n∈/ℓ∞ and (bn)n∈ℓ∞(b_n)_n\in\ell^\infty(bn)n∈ℓ∞D(an)n∉ℓ∞(a_n)_n\notin\ell^\infty(an)n∈/ℓ∞ and (bn)n∉ℓ∞(b_n)_n\notin\ell^\infty(bn)n∈/ℓ∞Pick an option to check your answer.Show answerShow solution Report a problem with this question← Q55All questionsQ57 →Keyboard: A–D answer · S solution · ←/→ previous/next