Proof

The theorem can be more rigorously stated as follows: ${\displaystyle \left(X\!\,-\!\,np\right)\!/\!{\sqrt {npq}}}$, with ${\displaystyle \textstyle X}$ a binomially distributed random variable, approaches the standard normal as ${\displaystyle n\!\to \!\infty }$, with the ratio of the probability mass of ${\displaystyle X}$ to the limiting normal density being 1. This can be shown for an arbitrary nonzero and finite point ${\displaystyle c}$. On the unscaled curve for ${\displaystyle X}$, this would be a point ${\displaystyle k}$ given by

${\displaystyle k=np+c{\sqrt {npq}}}$

For example, with ${\displaystyle c}$ at 3, ${\displaystyle k}$ stays 3 standard deviations from the mean in the unscaled curve.

The normal distribution with mean ${\displaystyle \mu }$ and standard deviation ${\displaystyle \sigma }$ is defined by the differential equation (DE)

${\displaystyle f'\!(x)\!=\!-\!\,{\frac {x-\mu }{\sigma ^{2}}}f(x)}$ with an initial condition set by the probability axiom ${\displaystyle \int _{-\infty }^{\infty }\!f(x)\,dx\!=\!1}$.

The binomial distribution limit approaches the normal if the binomial satisfies this DE. As the binomial is discrete the equation starts as a difference equation whose limit morphs to a DE. Difference equations use the discrete derivative, ${\displaystyle \textstyle p(k\!+\!1)\!-\!p(k)}$, the change for step size 1. As ${\displaystyle \textstyle n\!\to \!\infty }$, the discrete derivative becomes the continuous derivative. Hence the proof need show only that, for the unscaled binomial distribution,

${\displaystyle {\frac {f'\!(x)}{f\!(x)}}\!\cdot \!\left(-{\frac {\sigma ^{2}}{x-\mu }}\right)\!\to \!1}$ as ${\displaystyle n\!\to \!\infty }$.

The required result can be shown directly:

${\displaystyle {\begin{aligned}{\frac {f'\!(x)}{f\!(x)}}{\frac {npq}{np\!\,-\!\,k}}\!&={\frac {p\left(n,k+1\right)-p\left(n,k\right)}{p\left(n,k\right)}}{\frac {\sqrt {npq}}{-c}}\\&={\frac {np-k-q}{kq+q}}{\frac {\sqrt {npq}}{-c}}\\&={\frac {-c{\sqrt {npq}}-q}{npq+cq{\sqrt {npq}}+q}}{\frac {\sqrt {npq}}{-c}}\\&\to 1\end{aligned}}}$

The last holds because the term ${\displaystyle -cnpq}$ dominates both the denominator and the numerator as ${\displaystyle n\!\to \!\infty }$.

As ${\displaystyle \textstyle k}$ takes just integral values, the constant ${\displaystyle \textstyle c}$ is subject to a rounding error. However, the maximum of this error, ${\displaystyle \textstyle {0.5}/\!{\sqrt {npq}}}$, is a vanishing value.^[4]

Alternative proof

The proof consists of transforming the left-hand side (in the statement of the theorem) to the right-hand side by three approximations.

First, according to Stirling's formula, the factorial of a large number n can be replaced with the approximation

${\displaystyle n!\simeq n^{n}e^{-n}{\sqrt {2\pi n}}\qquad {\text{as }}n\to \infty .}$

Thus

${\displaystyle {\begin{aligned}{n \choose k}p^{k}q^{n-k}&={\frac {n!}{k!(n-k)!}}p^{k}q^{n-k}\\&\simeq {\frac {n^{n}e^{-n}{\sqrt {2\pi n}}}{k^{k}e^{-k}{\sqrt {2\pi k}}(n-k)^{n-k}e^{-(n-k)}{\sqrt {2\pi (n-k)}}}}p^{k}q^{n-k}\\&={\sqrt {\frac {n}{2\pi k\left(n-k\right)}}}{\frac {n^{n}}{k^{k}\left(n-k\right)^{n-k}}}p^{k}q^{n-k}\\&={\sqrt {\frac {n}{2\pi k\left(n-k\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {nq}{n-k}}\right)^{n-k}\end{aligned}}}$

Next, the approximation ${\displaystyle {\tfrac {k}{n}}\to p}$ is used to match the root above to the desired root on the right-hand side.

${\displaystyle {\begin{aligned}{n \choose k}p^{k}q^{n-k}&\simeq {\sqrt {\frac {1}{2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {nq}{n-k}}\right)^{n-k}\\&\simeq {\frac {1}{\sqrt {2\pi npq}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {nq}{n-k}}\right)^{n-k}\qquad p+q=1\\\end{aligned}}}$

Finally, the expression is rewritten as an exponential and the Taylor Series approximation for ln(1+x) is used:

${\displaystyle \ln \left(1+x\right)\simeq x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots }$

Then

${\displaystyle {\begin{aligned}{n \choose k}p^{k}q^{n-k}&\simeq {\frac {1}{\sqrt {2\pi npq}}}\exp \left\{\ln \left(\left({\frac {np}{k}}\right)^{k}\right)+\ln \left(\left({\frac {nq}{n-k}}\right)^{n-k}\right)\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{-k\ln \left({\frac {k}{np}}\right)+(k-n)\ln \left({\frac {n-k}{nq}}\right)\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{-k\ln \left({\frac {np+x{\sqrt {npq}}}{np}}\right)+(k-n)\ln \left({\frac {n-np-x{\sqrt {npq}}}{nq}}\right)\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{-k\ln \left({1+x{\sqrt {\frac {q}{np}}}}\right)+(k-n)\ln \left({1-x{\sqrt {\frac {p}{nq}}}}\right)\right\}\qquad p+q=1\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{-k\left({x{\sqrt {\frac {q}{np}}}}-{\frac {x^{2}q}{2np}}+\cdots \right)+(k-n)\left({-x{\sqrt {\frac {p}{nq}}}-{\frac {x^{2}p}{2nq}}}-\cdots \right)\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{\left(-np-x{\sqrt {npq}}\right)\left({x{\sqrt {\frac {q}{np}}}}-{\frac {x^{2}q}{2np}}+\cdots \right)+\left(np+x{\sqrt {npq}}-n\right)\left(-x{\sqrt {\frac {p}{nq}}}-{\frac {x^{2}p}{2nq}}-\cdots \right)\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{\left(-np-x{\sqrt {npq}}\right)\left(x{\sqrt {\frac {q}{np}}}-{\frac {x^{2}q}{2np}}+\cdots \right)-\left(nq-x{\sqrt {npq}}\right)\left(-x{\sqrt {\frac {p}{nq}}}-{\frac {x^{2}p}{2nq}}-\cdots \right)\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{\left(-x{\sqrt {npq}}+{\frac {1}{2}}x^{2}q-x^{2}q+\cdots \right)+\left(x{\sqrt {npq}}+{\frac {1}{2}}x^{2}p-x^{2}p-\cdots \right)\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{-{\frac {1}{2}}x^{2}q-{\frac {1}{2}}x^{2}p-\cdots \right\}\\&={\frac {1}{\sqrt {2\pi npq}}}\exp \left\{-{\frac {1}{2}}x^{2}(p+q)-\cdots \right\}\\&\simeq {\frac {1}{\sqrt {2\pi npq}}}\exp \left\{-{\frac {1}{2}}x^{2}\right\}\\&={\frac {1}{\sqrt {2\pi npq}}}e^{\frac {-(k-np)^{2}}{2npq}}\\\end{aligned}}}$

Each "${\displaystyle \simeq }$" in the above argument is a statement that two quantities are asymptotically equivalent as n increases, in the same sense as in the original statement of the theorem—i.e., that the ratio of each pair of quantities approaches 1 as n → ∞.