Log-Series Distribution

The terms in the series expansion of $\ln(1-\theta)$ about $\theta=0$ are proportional to this distribution.

$$P(n) = -{\theta^n\over n\ln(1-\theta)}$$ (1)

$$D(n) \equiv \sum_{i=1}^n P(i) = {\theta^{1+n}\Phi(\theta,1,1+n)+\ln(1-\theta)\over\ln(1-\theta)},$$ (2)

where $\Phi$ is the Lerch Transcendent. The Mean, Variance, Skewness, and Kurtosis

$$\mu = {\theta\over(\theta-1)\ln(1-\theta)}$$ (3)

$$\sigma^2 = -{\theta[\theta+\ln(1-\theta)]\over (\theta-1)^2[\ln(1-\theta)]^2}$$ (4)

$$\gamma_1 = {2\theta^2+3\theta\ln(1-\theta)+(1+\theta)\ln^2(1-\theta)\over \l... ...heta)[\theta+\ln(1-\theta)] \sqrt{-\theta[\theta+\ln(1-\theta)]}} \ln(1-\theta)$$ (5)

$$\gamma_2 = {6\theta^3+12\theta^2\ln(1-\theta)+\theta (7+4\theta)\ln^2(1-\the... ...]^2}+{(1+4\theta+\theta^2)\ln^3(1-\theta)\over \theta[\theta+\ln(1-\theta)]^2}.$$ (6)