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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.

The Watterson estimator isThe expectation above makes it an unbiased estimator. Its variance, given by the variance of the Watterson estimator, isIndeed, the harmonic sum satisfies $H_{n-1}\sim \log n$ and $H_{n-1}^{(2)}$ stays bounded. This proves mean-square convergence; the Chebyshev inequality then shows that it is a consistent estimator, since $P(|\hat{\theta }-\theta |>\epsilon )\le \mathrm{Var}(\hat{\theta })/{\epsilon }^2\to 0$.

Use $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and normalized measure $dt/(2\pi )$. The Fourier partial sum at zero is the real linear functionalwhere the Dirichlet kernel isThe quotient follows by summing the finite geometric progression, with its removable value $D_N(0)=2N+1$. It is a real continuous periodic function. Ifthen $|L_{N}f|\le {\Lambda }_N\Vert f{\Vert }_{\infty }$. This bound is its exact operator norm even on the real continuous functions. Indeed, for $\delta >0$ setIt is real, continuous and periodic with norm at most one, andby dominated convergence. At zeros of the kernel the integrand is zero, and elsewhere its limit is $|D_N|$.

For $0<t\le \pi$, $\sin (t/2)\le t/2$. Using evenness and putting $v=(N+1/2)t$ givesFor the last inequality, split into the first $N$ complete intervals $[(j-1)\pi ,j\pi ]$: their sine integrals are two and $1/v\ge 1/(j\pi )$. The harmonic sum tends to infinity. This proves the Dirichlet kernel harmonic lower bound.

Choose $N$ with ${\Lambda }_N>K$, then choose $\delta$ sufficiently small that $L_N{f}_{N,\delta }>K$. ThusThis constructs continuous near-maximizers, rather than using the discontinuous sign of the kernel as the desired function.

Expand the two Dirichlet polynomials, keeping the complex conjugate on $B$. The diagonal $m=n$ contributes exactly $T{\sum }_{n\le X}{a}_n\overline{b_n}/n^{2\sigma }$. If $m\ne n$, the exponential integral iswhose absolute value is at most $2/|\log (m/n)|$. Thus the mean-value formula isThis also covers $T=0$.

For $m>n$, integration of $1/x$ over $[n,m]$ gives $\log (m/n)\ge (m-n)/m$; the analogous inequality holds with $m,n$ exchanged. Hence $|\log (m/n){|}^{-1}\le \max (m,n)/|m-n|\le X$. This gives the requested first bound $O(X{\sum }_{m\ne n}|a_n||b_m|/(n^{\sigma }{m}^{\sigma }))$.

For the weighted bound, put ${\alpha }_n=|a_n|n^{-\sigma }$ and ${\beta }_m=|b_m|m^{-\sigma }$. The inequalityreduces the estimate to a row sum and its symmetric counterpart. For a fixed $n$, the preceding logarithm bounds giveHere $n^2/(m(n-m))=n(1/m+1/(n-m))$, and each resulting harmonic sum is $O(\log (2X))$. The symmetric row sum is bounded by $O(m\log (2X))$. Therefore the second error bound is

Expand the finite Euler product into geometric series. By unique prime factorization, it is the sum of $1/n$ over positive integers whose prime factors do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The harmonic sum satisfiesThe last inequality uses $m+1>x$, including when $x$ is an integer. For each prime number $p$,The telescoping series over all integers $n\ge 2$ bounds the sum of these remainders by $\frac{1}{2}{\sum }_{n\ge 2}[n(n-1){]}^{-1}=1/2$. Taking logarithms of the product bound therefore givesThis explicit constant suffices; no asymptotic theorem about the distribution of prime numbers is needed.

Writeuniformly for $j\ge 1$. The accumulated error from $j=n+1$ to $2n$ tends to zero, while the corresponding harmonic sum tends to $\log 2$. Hence $b_n\to \log 2$.