I am interested in finding any upper bound for the following series:
$$\displaystyle f(x) := \sum_{n \in \mathbb{Z}^d \backslash \{0\}} |n|^{-d/2}J_{d/2}(k|n|)e^{inx},$$
where $d \geqslant 2$, $x \in \mathbb{R}^d$, $k > 0$ is a positive constant, and $J_{\nu}$ is the Bessel function of the first kind. (For background, this sum arises in the study of the $L^p$-norms of the remainder in the (generalised) Gauss circle problem.) The sight of the exponential made me immediately want to take absolute values. However, using the bound $|J_{\nu}(z)| \leqslant C_{\nu}|z|^{-1/2}$, which comes from this for large $z > 0$, then we get:
$$\displaystyle |f(x)| \leqslant C_d \sum_{n \in \mathbb{Z}^d \backslash \{0\}}|n|^{-\frac{d+1}{2}},$$
and this series diverges, since it would converge only if $d < 1$, which is impossible (see here for the details). So bounding via modulus does not seem fruitful. Therefore, it seems more logical to exploit the positive-negative cancellation of the terms of the Bessel function. We could use the asymptotic expansion:
$$ J_\nu (x) \sim \sqrt{\frac{2}{\pi x}} \cos \left( x - \frac{2 \nu +1}{4} \pi \right) + O \left( \frac{1}{x^{3/2}} \right), \;\;\;\;x \to \infty,$$
but then the $O(x^{-3/2})$ term forces us to take $d = 2$ rather than get a sum which converges for general $d$. Does anyone have any better ideas for how we could find an upper bound for this series? Are there any known results for similar sums?