This a follow-up to the question: Convergence Rate of the Tail of the Series $m^{a}\sum_{n=1}^{\infty}\min\{1,2^n m^{-1} \}2^{-na}$
When $a > 0$, we have $$ \sum_{n=1}^{\infty}\min\{1,2^n |x|^{-1} \}2^{-na} \leq C|x|^{-a} \quad \forall x \in \mathbb{R}, x\neq 0. $$ where $C$ is a finite constant independent of $x$. Then for each $N$ $$ \sum_{n=N}^{\infty}\min\{1,2^n |x|^{-1} \}2^{-na} \leq C|x|^{-a} \quad \forall x \in \mathbb{R}, x\neq 0. $$ and for each fixed $x$ $$ \sum_{n=N}^{\infty}\min\{1,2^n |x|^{-1} \}2^{-na} \rightarrow 0 \quad \text{as} \quad N \rightarrow \infty. $$ The dominated convergence theorem implies $$ \left\| \sum_{n=N}^{\infty}\min\{1,2^n |x|^{-1} \}2^{-na} \right\|_{L^p} \rightarrow 0 \quad \text{as} \quad N \rightarrow \infty. $$ when $p > 1/a$. I want to know how fast the tail converges in $L^p$. I'd like to get a bound like $$ 2^{N(a+b)}\left\| \sum_{n=N}^{\infty}\min\{1,2^n |x|^{-1} \}2^{-na} \right\|_{L^p} \rightarrow 0 \quad \text{as} \quad N \rightarrow \infty. $$ for some $b>0$. I am most interested in the case when $1/a < p < 1/a + \epsilon$, where $\epsilon$ is a small.
I don't think this question is trivial, but if it is please tell me how.