Contradiction to $2^n \ge c\cdot 6^\frac{n}{2}$

Question

I want to prove that the statement $2^n \in \Omega(6^\frac{n}{2})$ is not true.

Suppose in a way of contradiction that there exists a $c>0$, and $n_0 \in \mathbb{N}$ such that for every $n>n_0$ we have $0 \le c\cdot 6^\frac{n}{2} \le 2^n$.

But for every $n>2c$ we have: $6^\frac{n}{2} > 6^c = (2^c)^{\log_26}=2^{c~log_26}>2^{2c}$.

But I do not know what to do with $c\cdot 6^\frac{n}{2}$...

Any help?

Thanks!

Answer 1

Hint

$$2^n\ge c\cdot 6^{n/2}\Leftrightarrow (2/3)^{n/2}\ge c.$$

See that $2/3<1$. So $(2/3)^{n/2}$ decreases and goes to zero, when $n$ increases.