Question: For what values of a does the sequence {n!} grow faster than the sequence {n an }? Stirlings formula is useful: n! 2pn n

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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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 full answer

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