algorithm - How do I prove all constants to some exponential power belong to little-o of some function

Question

I'm trying to prove that c²ⁿ = o((loglog n)ⁿ) (That's little-o) for any constant c. I understand that we can prove one function grows at a smaller rate than the other by taking the limit as n approaches infinity, and I can very easily pick some arbitrary integer value for c and show that indeed ((loglog n)ⁿ) grows at a faster rate. But how do I prove this to be true for any constant c?

Answer 1

你想证明对于任何 c 的选择，

lim _{n → ∞} (c ²ⁿ / (log log n) ⁿ ) = 0

请注意，对于 c 的任何选择，

lim _{n → ∞} (c ²ⁿ / (log log n) ⁿ )

= lim _{n → ∞} (c ² / log log n) ⁿ

此限制为 0，因为一旦 log log n > c ²，您将提高小于 1 的 n 次方的值，该值将迅速降至零。

希望这可以帮助！