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?
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Some times, you are just too stressful, so you think you have problem proving the bound when you replace the constant $c$ with another constant $d$. But you don't have any problem with the constant $d$ at all, it's just some minor issue that will not affect the proof as a whole. – InformedA Sep 02 '14 at 07:34
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Try proving that $5^{2n} = o((\lg \lg n)^n)$. Write out the proof in detail, and make sure you understand every step.
Then, try replacing the $5$ in the base with an arbitrary constant $c$. Does every step still go through? Hint: you should be able to extend the argument to handle an arbitrary constant $c$, with very little change.
If you're still stuck, edit the question to show us what you've tried, including showing us your proof that $5^{2n} = o((\lg \lg n)^n)$ and your reasoning about which steps are still valid when you replace $5$ with an arbitrary constant $c$.
D.W.
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