choice of $n_1$ and $n_2$ {#sec2.4}
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In practice it is not difficult to choose $n_1$ and $n_2$ so that (\[eq2.15\]) holds. To see how, assume for simplicity that the lattice edge widths $e_1$ and $e_2$ are identical, as they would be if (for example) $S$ were equal to a sum or difference of estimators of binomial proportions. If $\rho_0$ is an irrational number then the convergents $m_1/m_2$ of $\rho_0$ satisfy $$\label{eq2.18} \bigl|(m_1/m_2)-\rho_0\bigr|\leq
m_2^{-2}.$$ (See e.g. Leveque [@Lev56], equation (29), p. 180.) Therefore, if $n_1$ and $n_2$ are relatively prime and $n_1/n_2$ is a convergent of $\rho_0$, then (\[eq2.15\]), for each $\delta\in(0,3-(1/\eta)]$, follows from (\[eq2.18\]). The most difficult case, as far as (\[eq2.15\]) is concerned, is the one where the convergence rate in (\[eq2.15\]) is fastest, and arises when $\eta=1$. There we need to ensure that $$\label{eq2.19} |\rho_{j_1j_2}-\rho_0|=O \bigl(n^{-1-\delta}
\bigr)$$ for some $\delta>0$. Now, (\[eq2.19\]) holds whenever $n_1/n_2$ is a convergent of $\rho_0$, and the Khinchin-Lévy Theorem (see, e.g., pp. 82–83 of Einsiedler and Ward [@Ein11]) implies that the convergents are reasonably closely spaced; the numerators and denominators generally increase by factors of only $\pi^2/(12 \log
2)\approx1.87$. Moreover, there are many ratios $n_1/n_2$ on either side of convergents for which 