For a bounded sequence $(a_k)_{k=1}^\infty$ define the sequence $(A_j)_{j=1}^\infty$ by$$A_j = \sum_{k=1}^\infty \frac{a_k}{k^2+j^4}.$$It is elementary to check that (I cheated and used Mathematica)$$|A_j| \leq \|a\|_\infty \sum_{k=1}^\infty \frac{1}{k^2+j^4} = \|a\|_\infty \frac{-1+j^2 \pi \operatorname{Coth}(j^2 \pi)}{2j^4},$$so that$$A_j = O\left( \frac{1}{j^2} \right).$$This bound is surely not sharp, and with the oscillating sequence $a_k = (-1)^k$ we obtain$$A_j \sim \frac{-1}{2j^4} = \Theta\left( \frac{1}{j^{4}} \right).$$However I am not able to go beyond $1/j^4$ (with a non-identically zero sequence).
Question: Can one find a non-identically zero sequence $(a_k)_{k=1}^\infty$ so that$$A_j = O\left( \frac{1}{j^{4+\epsilon}} \right),$$for some $\epsilon > 0$?
Trying with $a_k = (-1)^k k^{-n}$ I obtain that$$A_j = \Theta\left( \frac{1}{j^{4}} \right).$$For $a_k = (-1)^ke^{-k}$ I obtain$$A_j = \frac{-1}{2ej^4}(\alpha_j + \beta_j) + \frac{i}{2ej^6}(\alpha_j-\beta_j)$$where the complex numbers $\alpha_j$ and $\beta_j$ are defined by$$\alpha_j = {}_2F_1(1,1-ij^2,2-ij^2,-1/e),\quad \beta_j = {}_2F_1(1,1+ij^2,2+ij^2,-1/e) .$$The function ${}_2F_1$ is the hypergeometric function, and it can be shown that the sequences $(\alpha_j)$ and $(\beta_j)$ are bounded. However it is not clear to me whether $\alpha_j + \beta_j$ is zero or not.