For what values of a does the sequence {n!} grow faster than the sequence {n an }?

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For what values of a does the sequence {n!} grow faster than the sequence {n^an}? Stirling’s formula is useful: n! ≈ √2pn nⁿe^-n, for large values of n.

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Calculus Early Transcendentals

ISBN: 978-0321947345

2nd edition

Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Question Posted: Jun 14, 2022 02:22 AM

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For what values of a does the sequence {n!} grow faster than the sequence {n an }?

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First note that for a 1 we already know that n grows fast than n So if a 1 then nan n so that nan ...View the full answer

Calculus Early Transcendentals

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