# Why does $k \to np$?

There's a section called Proof, where k is defined as $$k=np + \sqrt{npq}x$$. Then it says that "From this definition we have the approximations $$k \to np$$" when $$n \to \infty$$ and I get stuck here. I don't understand why $$k \to np$$.

• As $n \to \infty$, the square root $\sqrt{n}$ becomes negligible compared to the linear term. – Lee David Chung Lin May 2 at 2:24
• It's a sloppiness on the page you cite. They write $k\to np$ but mean $k/n\to p$ or $k\sim np$. Both $k$ and $np$ tend to infinity in the DeMoivre-Laplace theorem, and, strictly speaking $k\to np$ is a no-no. But one that good mathematicians use informally. – kimchi lover May 2 at 2:28

Note that $$np+(\sqrt{npq})x=np\left(1+\frac{(\sqrt{npq})x}{np}\right)=np(1+O(1/\sqrt n))$$. Thus, as $$n\to\infty$$, we have $$k\sim np$$. In particular, all the author of the article means is that $$k\to\infty$$ at exactly the same rate as $$np$$ as $$n\to\infty$$. These types of arguments are particularly useful when needing to work with asymptotics rather than exact values; it gives us an easy way to bound quantities.
• Thank you for your answer. But $np+x\sqrt{np}=\sqrt{np}(\sqrt{np}+x)$ and if you take the limit when $n \to \infty$, then you get $\infty * \infty = \infty$. What do you think about it? – roy212 May 2 at 2:32
• @roy212: Yes? $k/np\to 1$ as $n\to\infty$. This means $k\approx np$ or more specifically, $k\sim np$. Both tend to infinity with $n$, and $k$'s rate of convergence to $\infty$ is exactly the same as $np$. – Clayton May 2 at 2:34
• If both tend to infinity. Why is it wrong when I say that "$np+x\sqrt{np}=\sqrt{np}(\sqrt{np}+x)$ and if you take the limit when $n \to \infty$, then you get $\infty * \infty = \infty$"? – roy212 May 2 at 2:41
• @roy212: it isn't wrong, it just doesn't provide as much information. As the comment underneath your post and my answer both state, $k\to\infty$ as $n\to\infty$ and $np\to\infty$ as $n\to\infty$. What is relevant is that $\frac{k}{n}\to p$ as $n\to\infty$. This contains very important information about how $k$ behaves for large values of $n$. What you have written only says that $k\to\infty$ but doesn't give a rate of convergence (or divergence, if you prefer to think of it that way). – Clayton May 2 at 2:43
• @roy212: Beyond my answer, I don't think so... clearly $1/\sqrt{n}\to0$ as $n\to\infty$. This leaves a $p\cdot(1)=p$ on the right-hand side. – Clayton May 2 at 2:47