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For fixed $\theta >0$, a geometric mutation count in a coalescent epoch with mean ${\mu }_i=\theta /i$ has third moment ${\mu }_i+6{\mu }_i^2+6{\mu }_i^3$. The centered absolute third moment is at most $4\mathbb{E}[Y_i^3]+4{\mu }_i^3$, so its sum is $O(\log n)$. The total variance is asymptotic to $\theta \log n$, making the Lyapunov condition with $\delta =1$ hold. Rescaling by the harmonic sum gives the displayed limit. At $\theta =0$ the estimator is identically zero and the limit is degenerate. The variance of the Watterson estimator decreases only logarithmically with sample size at a fixed nonrecombining locus.

For independent centered random variables with total variance $s_n^2$, the Lyapunov condition requires the displayed expression to tend to zero for some $\delta >0$. The inequality $z^2{1}_{\{|z|>\epsilon s_n\}}\le |z{|}^{2+\delta }/(\epsilon s_n{)}^{\delta }$ implies the Lindeberg condition after normalization. The Lindeberg-Feller central limit theorem then makes the normalized sum converge to the standard normal distribution.

For fixed $\theta >0$, use the Lyapunov condition for the independent centered epoch counts $Z_i=Y_{i+1}-\theta /i$. Their total variance is $s_n^2=\theta H_{n-1}+{\theta }^2{H}_{n-1}^{(2)}\sim \theta \log n$.

Write ${\mu }_i=\theta /i$. Differentiating the geometric distribution probability generating function gives $\mathbb{E}[Y_{i+1}^3]={\mu }_i+6{\mu }_i^2+6{\mu }_i^3$. Since $|Y_{i+1}-{\mu }_i{|}^3\le 4Y_{i+1}^3+4{\mu }_i^3$, the sum of the centered absolute third moments is $O(\log n)$. ConsequentlyThe Lindeberg-Feller central limit theorem applies through the Lyapunov condition, giving $(S-\theta H_{n-1})/s_n\stackrel{d}{\to }N(0,1)$. Now $\sqrt{\log n}{s}_n/H_{n-1}\to \sqrt{\theta }$, so the Slutsky theorem proves the asymptotic normality of the Watterson estimator:The limiting variance is $\theta$. When $\theta =0$, every count vanishes and the limiting law is the point mass at zero.