integers into adjacent blocks each comprised of $2 \lfloor n^{\alpha}\rfloor+1$ consecutive integers. Write ${\bar\nu}_\ell$ for the central integer in the $\ell$th block, which we denote by ${{\mathcal N}}_\ell$ where $-\infty<\ell<\infty$ and ${{\mathcal N}}_{\ell+1}$ is immediately to the right of ${{\mathcal N}}_\ell$ on the number line. Given $\nu\in{{\mathcal N}}_\ell$, put $\nu_\ell=\nu-{\bar\nu}_\ell$.

Let $c_3=e_2 n_1/{\sigma}_1 n_2$ and $c_4=(e_1/{\sigma}_2) (n_1/n_2)^{1/2}$, and note that $c_1,\ldots,c_4$ are strictly positive functions of $n$ and are bounded away from zero and infinity as $n$ diverges. Put ${\gamma}=\prod_{j=1,2} (e_j/{\sigma}_j)$, and, given an integer $r_0\geq1$, define $$\begin{aligned}
\label{eq2.10} \phi(u,x)&=&\phi \bigl\{(x/c_1)-c_3 u
\bigr\} \phi(c_4 u), \qquad\phi_r(u,x)=({\partial}/{\partial}u)^r \phi(u,x),
\nonumber
\\
K_n(x)&=&{\gamma}\sum_{r=0}^{r_0}
\sum_{-\infty<\ell<\infty} \frac
{\phi_r ({\bar\nu}_\ell /n_1^{1/2},x )}{ r! n_1^{r/2}} \sum
_{\nu\in{{\mathcal N}}_\ell} \nu_\ell^r \psi \biggl\{
\xi_n(x)-\frac{e_2 n_1}{
e_1 n_2} \nu \biggr\},\end{aligned}$$ where, as in Section \[sec1.1\], $\psi(x)=\lfloor x\rfloor
-x+{\frac{1}{2}}$.

We claim that the infinite series in the definition of $K_n(x)$ is absolutely convergent, uniformly in $x$. To appreciate why, note that $$\label{eq2.11} \sup_{-\infty<x<\infty} \bigl|\phi_r \bigl({\bar
\nu}_\ell /n_1^{1/2},x \bigr) \bigr| \leq
C_1(r) \phi \bigl(c_4 {\bar\nu}_\ell
/n_1^{1/2} \bigr),$$ where, here and below, the notation $C_j(r)$ will denote a constant depending on $r$ but not on $n$. Using (\[eq2.4\]) and (\[eq2.11\]), we deduce that $$\label{eq2.12} \sum_{-\infty<\ell<\infty} \Bigl\{\sup_{-\infty<x<\infty}
\bigl|\phi_r \bigl({\bar\nu}_\ell /n_1^{1/2},x
\bigr) \bigr| \Bigr\} \leq 