n, n \pm 1, \label{eq:zbd}$$ and if $\vert n - j \vert \geq 2$, $$\frac{3}{2} \vert n - j \vert \leq 2 \vert n - j \vert - 1 \leq \vert z - z_j \vert \leq 2 \vert n - j \vert + 1 \leq \frac{5}{2} \vert n - j \vert.  \label{eq:njest}$$ Therefore, by , , $$\ell^2 = \sum_{j, k = 1}^{\infty} \frac{\vert w_{jk} \vert^2}{\vert z - z_j \vert \vert z - z_k \vert} \leq C_0^2 \mu^2 \label{eq:nuexpr}$$ with $$\mu = \sum_{j = 0}^{\infty} \frac{1}{(1 + j)^{2 \alpha} \vert z - z_j \vert}. \label{eq:mudef}$$ The sum of three terms for $j = n, n \pm 1$ in by does not exceed $$3 \cdot \frac{1}{n^{2\alpha}} \cdot 2 = \frac{6}{n^{2\alpha}},  \label{eq:smallvarianceterms}$$ and by , the remaining part of $\mu$, namely, $ {\displaystyle}\sum_{j = 0}^{n - 2} + \sum_{j = n + 2}^{\infty} $, by the integral test does not exceed $$\begin{aligned}
&\frac{2}{3} \left[ \frac{1}{n} + \frac{1}{n^{2\alpha}} + \int_0^{n - 1} \frac{dx}{(1 + x)^{2\alpha} (n - x)} \right]\\
+ & \frac{2}{3} \left[ \frac{1}{2} \cdot \left( \frac{1}{n + 3} \right)^{2\alpha} + \int_{n + 2}^{\infty} \frac{dy}{(1 + y)^{2\alpha} (y - n)} \right].
\end{aligned} \label{eq:bounda}$$ The first integral (after the change of variables $x = n 