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Let $n\geq 1, m>1, k\leq n$.

I am trying to find condition on $m,$ that $$ 4\sqrt{\pi}(2m)^{mn}\leq2^k $$

Thank you.

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1 Answer 1

up vote 2 down vote accepted

$4\sqrt{\pi}(2m)^{mn} = 4\sqrt{\pi}2^{mn}m^{mn} \geq 4\sqrt{\pi}2^k m^{mn}$ since $k \leq n$ and $m > 1$.

Clearly, $4\sqrt{\pi}2^k m^{mn} > 2^k$ for all $k \in \mathbb{N}$.

Are you sure that the conditions on $n,m$ and $k$ are correct ?

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