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How to prove non-equivalence of this?

$a^n = \Omega(b^n)$

when

$0 < a < b$

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1 Answer 1

up vote 1 down vote accepted

Suppose $a^n=\Omega(b^n)$ i.e. $a^n\ge c\cdot b^n$ for some $c>0$ and all $n>N$ for some $N$. Now $$0<a<b \implies 0<a/b<1 \implies \log(a/b)<0$$ $$\implies n\log(a/b)\to-\infty\implies\exp(*)=(a/b)^n\to0$$ which implies for any $\epsilon>0$ there is an $M$ such that $n>M\implies(a/b)^n<\epsilon$. Specialize $\epsilon=c$ and set $k=\max\{N,M\}+1$ in which case $a^k\ge c\cdot b^k$ (by hypothesis) and $(a/b)^k<c$ (by deduction), a contradiction, hence $a^n\ne\Omega(b^n)$.

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