In my previous question, I tried applying the Residue Theorem to compute the sum $$S:= \sum_{n=2}^{\infty} \frac{ \operatorname{sinc} (-4\pi (n-3)(n-4) ) }{n(n-1)} ,$$ where $\operatorname{sinc}(\cdot)$ is the normalized sinc function.
In general, sums can be evaluated by means of this theorem according to the following result: $$\lim_{k \to +\infty} \sum_{k=-N}^{N} f(k) = - \{ \text{sum of the residues of } \pi f(z) \cot(\pi z) \text{ at the poles of }f(z) \} . \qquad \qquad (*)$$
In this case, the poles of $f(z):= \frac{ \operatorname{sinc} (-4\pi (z-3)(z-4) ) }{z(z-1)} $ are there at $z=0$ and $z=1$, and the residues of $\pi f(z) \cot(\pi z) $ at these values amount to $\frac{7}{12}$ and $ -\frac{10}{12}$. Thus, we obtain
\begin{align} S: &= \sum_{n=-\infty}^{\infty} \frac{ \operatorname{sinc} (-4\pi (n-3)(n-4) ) }{n(n-1)} \\ &= \sum_{n=0}^{\infty} \frac{ \operatorname{sinc} (-4\pi (n-3)(n-4) ) }{n(n-1)} \\ &= - \Big{(} \frac{7}{12} - \frac{10}{12} \Big{)} \\ &= \frac{1}{4}. \end{align}
This is the correct answer. However, as user metamorphy pointed out in the comments to the previous question, the Residue Theorem as described in $(*)$ may not be applied in this case. This is due to the fact that $f$ must satisfy the condition that $$|f(z)| < \frac{M}{z^{k}} $$ - where $k>1$ and $M$ are constants independent of $N$ - along the path $C_{N}$. This path goes counterclockwise along the rectangle $(N + (1/2))(1-i)$, $(N+(1/2))(1+i)$, $(N+(1/2))(-1-i)$ and $(N+(1/2))(-1-i)$ and it encloses the poles of $f$. See p. 3-5 of this document for more details.
So my question boils down to whether we could somehow tweak $f$ in such a way that the Residue Theorem can be applied, and that the same value for $S$ can be obtained. Are there any tricks to adjust $f$ in such a way that the Residue Theorem may be used to obtain the correct value of $S$ ?