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I am trying to evaluate the combinatorics term of (n choose k) where k is not constant but $k=o\left(\sqrt{n}\right)$ (as in, small-o and not big-O). Following the suggestion from here n-choose-k-theta

I have reached out to :

$$\frac{1}{k!}\left(n\left(n-1\right)\left(n-2\right)...\left(n-k+1\right)\right)=O\left(\frac{1}{k!}n^{k}\right)$$

And:

$$\frac{1}{k!}\left(n\left(n-1\right)\left(n-2\right)...\left(n-k+1\right)\right)=\frac{1}{k!}n^{k}\left(1\cdot\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{k}{n}\right)\right)=\frac{1}{k!}n^{k}\Omega\left(\left(1-\frac{k}{n}\right)^{k}\right)$$

But now I am stuck to show that for $k=o\left(\sqrt{n}\right)$ : $$\Omega\left(\left(1-\frac{k}{n}\right)^{k}\right)=\Omega\left(1\right)$$

Which doesn't seem very natural to me at least.

linuxbeginner
  • 67

1 Answers1

4

We have that

$$\left(1-\frac{k}{n}\right)^{k}=e^{k\log\left(1-\frac{k}{n}\right)}$$

and since by definition

$$k=o(\sqrt{n})=\omega(n) \sqrt n$$

with $\omega (n) \to 0$, we have

$$k\log\left(1-\frac{k}{n}\right)=\omega(n) \sqrt n\log\left(1-\frac{\omega(n)\sqrt n}{n}\right)\sim \omega(n) \sqrt n\left(-\frac{\omega(n)\sqrt n}{n}\right)=-(\omega(n))^2\to 0$$

and thus

$$\left(1-\frac{k}{n}\right)^{k}\sim 1$$

user
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