For large value of n, and moderate values of the probabiliy of success p (roughly, 0.05 Leftarrow p Leftarrow 0.95), the binomial distribution can be

Step 1
Alternatively prove that:
Theorem:
If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the
variablez $=\frac{x-\mu }{\sigma }-N(0,1)$
where $\mu =np{\sigma }^2=np(1-p)$
Proof:
It can be prove using Moment generating function for binomial distribution. It's given as,
$M_x(\theta )=(q+p{e}^{\theta }{)}^n$
where $q=1-p.$
Step 2
By the linear transformation properties of the moment generating function.
$M_z(\theta )=e-\frac{\mu \theta }{\sigma }{M}_x\left(\frac{\theta }{\sigma }\right)=e-\frac{\mu \theta }{\sigma }{(q+pe\frac{\theta }{\sigma })}^n$
Taking the natural log of both sides, and then expanding the power series of $e^{\theta /\sigma }$
Then, $\ln M_z(\theta )=-\frac{\mu \theta }{\sigma }+n\ln (q+pe\frac{\theta }{\sigma })=-\frac{\mu \theta }{\sigma }+n\ln (q+p\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k!}{\left(\frac{\theta }{\sigma }\right)}^k)$
Since $p+q=1.$
$=-\frac{\mu \theta }{\sigma }+n\ln (1+p\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k!}{\left(\frac{\theta }{\sigma }\right)}^k)$
If n is made sufficiently large $\sigma =\sqrt{npq}$ can be made large enough that for any fixed theta the absolute value of the sum above will be less than 1.
Let $t=p\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m!}{\left(\frac{\theta }{\sigma }\right)}^m$
Thus for sufficiently large $n,|t|<1.$
The ln term in the previous expression is $\ln (1+t)where|t|<1,$ and so we may expand this term as follows:
$\ln (1+t)=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k-1}\frac{t^k}{k}$
Step 3
This means that
$\ln M_z(\theta )=-\frac{\mu \theta }{\sigma }+n\ln (1+t)=-\frac{\mu \theta }{\sigma }+n\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k_1}$

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