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I am trying to understand the DeMoivre-Laplace Theorem in this page.

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 $k \to np$.

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  • $\begingroup$ As $n \to \infty$, the square root $\sqrt{n}$ becomes negligible compared to the linear term. $\endgroup$ – Lee David Chung Lin May 2 at 2:24
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    $\begingroup$ 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. $\endgroup$ – kimchi lover May 2 at 2:28
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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.

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  • $\begingroup$ 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? $\endgroup$ – roy212 May 2 at 2:32
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    $\begingroup$ @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 $np$. $\endgroup$ – Clayton May 2 at 2:34
  • $\begingroup$ 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 $\infty * \infty = \infty$"? $\endgroup$ – roy212 May 2 at 2:41
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    $\begingroup$ @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). $\endgroup$ – Clayton May 2 at 2:43
  • $\begingroup$ @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. $\endgroup$ – Clayton May 2 at 2:47

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