For $m = m(n) \to \infty, {n \choose 2} - m \to \infty, N = {n \choose 2}$, we have $(1 + o(1))\sqrt{\frac{N}{2\pi m(N-m)}} \geq \frac{1}{10\sqrt{m}}$

Question

This is a step in a proof that I’m reading. How do we get this bound?

Answer 1

We divide both sides by $\sqrt{\frac{N}{2\pi m (N-m)}}$ and simplify to see that it suffices to show

$$\sqrt{\frac{2 \pi}{100}} \cdot \sqrt{1 - \frac{m}{N}} \leq 1 + o(1).$$

Now as long as $m = o(N)$, we know that

$$\sqrt{1 - \frac{m}{N}} \leq 1 + \frac{m}{2N} = 1 + o(1).$$

But $\sqrt{\frac{\pi}{50}} \approx 0.25 < 1$. In particular, this gives us

$$ \sqrt{\frac{\pi}{50}} \cdot \sqrt{1-\frac{m}{N}} \leq 0.251(1 + o(1)) \leq 1 + o(1) $$

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I hope this helps ^_^