Kurtosis and Skewness of Binomial Distribution Let $X \thicksim B(n,p)$ then I would like to evaluate kurtosis and skewness of X.
First I want to use the fact that kurtosis $k_3(\dfrac{X-\mu}{ \sigma})=\dfrac{k_3(X)}{\sigma^3}$ and skewness kurtosis $k_4(\dfrac{X-\mu}{ \sigma})=\dfrac{k_4(X)}{\sigma^4}$.
To use above identity, one needs to derive 3 and 4-th cumulant of X.
mgf of bionomial $X$ is $ M(t)=[(1-p)+pe^t]^n$ thus 
$K(t) = \log M(t)=n\log [(1-p)+pe^t]$
My question is here: 
How could one expand above log term into the form such as $\sum_{r=1}^{\infty}\dfrac{k_r(0)}{r!}t^r$?
textbook has denoted 
$n\log [(1-p)+pe^t] = \sum_{r=1}^{\infty}\dfrac{t^r}{r!}\{n\sum_{k=1}^r\sum_{j_1+j_2...+j_k=r}\begin{pmatrix}r\\j_!,j_2...j_k\end{pmatrix}\dfrac{(-1)^{k-1}}{k}p^k\}$
How could above identity be derived?
 A: To calculate the derivatives up to the 4th you can do them by hand and make sure you don't make any errors. To do this you'll need to use chain rule, quotient rule, product rule, and lots of organization and notebook paper. Or you can use an online differentiator-this is what I did using this one (http://www.derivative-calculator.net/). Where you can enter the function as
$$nlog(1-p+pe^x)$$ 
because $x$ is the variable it is programmed to differentiate with respect to and you can get all 4 derivatives. Writing the function in the OP as $f(t)$ they are: 
$$f^\prime(t)=\frac{npe^t}{1-p+pe^t}$$
$$f^{(2)}(t)=\frac{np(1-p)e^t}{(1-p+pe^t)^2}$$
$$f^{(3)}(t)=\frac{np(p-1)(1-p+pe^t)e^t}{(1-p+pe^t)^3}$$
$$f^{(4)}(t)=\frac{n(1-p)pe^t(p^2e^{2t}+(4p^2-4p)e^t +p^2-2p+1)}{(1-p+pe^t)^4}$$
now evaluating each of these at $t=0$ gives the first 4 cumulants, denoted $k_i$ for $i=1,2,3,4$. These cumulants are: 
$$k_1=np$$
$$k_2=np(1-p)$$
$$k_3=-np(1-p)(2p-1)$$
$$k_4=np(1-p)(6p(p-1)+1)$$
and you can check that the skewness is: 
$$\frac{k_3}{k_2^{3/2}}=\frac{-np(1-p)(2p-1)}{np(1-p)\sqrt{np(1-p)} }=\frac{-(2p-1)}{\sqrt{np(1-p)}}$$
and the kurtosis is: 
$$\frac{k_4}{k_2^{2}}=\frac{np(1-p)(6p(p-1)+1)}{n^2p^2(1-p)^2}=\frac{6p(p-1)+1}{np(1-p)}=\frac{1-6p(1-p)}{np(1-p)}$$
The expression you have written for the infinite series is the taylor series expansion of the function $f(t)$. 
